Proof of exponential property
$begingroup$
Given $f(x) = exp(ax)$, then the derivative, by definition is given by $frac{df}{dx} = aexp(ax)$. Also, it is known that $f(0) = 1$
I want to prove that $exp((a+b)x)$ is equal to $exp(ax)cdotexp(bx)$.
So, to do this, I take the derivative of $exp((a+b)x)$ which results in $(a+b)exp((a+b)x)$ by definition of the exponential. Next, I take the derivative of $exp(ax)cdotexp(bx)$ which results in $(a+b)exp(ax)cdotexp(bx)$.
Now, I evaluate $frac{d}{dx}$$[exp((a+b)x) - exp(ax)cdotexp(bx)]$ at $x = 0$. This results in zero.
My question now is, have I completed the proof. If yes, can someone explain why?
calculus exponential-function
edited Dec 6 '18 at 5:23
Tianlalu
3,08421138
asked Dec 6 '18 at 1:05
GustoCoGustoCo
428
$endgroup$
add a comment |
$begingroup$
Given $f(x) = exp(ax)$, then the derivative, by definition is given by $frac{df}{dx} = aexp(ax)$. Also, it is known that $f(0) = 1$
I want to prove that $exp((a+b)x)$ is equal to $exp(ax)cdotexp(bx)$.
So, to do this, I take the derivative of $exp((a+b)x)$ which results in $(a+b)exp((a+b)x)$ by definition of the exponential. Next, I take the derivative of $exp(ax)cdotexp(bx)$ which results in $(a+b)exp(ax)cdotexp(bx)$.
Now, I evaluate $frac{d}{dx}$$[exp((a+b)x) - exp(ax)cdotexp(bx)]$ at $x = 0$. This results in zero.
My question now is, have I completed the proof. If yes, can someone explain why?
calculus exponential-function
edited Dec 6 '18 at 5:23
Tianlalu
3,08421138
asked Dec 6 '18 at 1:05
GustoCoGustoCo
428
$endgroup$
3
$begingroup$
No. You have two functions $f$ and $g$, that you differentiate, and you check the derivatives are equal at $0$. Does it mean $f$ and $g$ are equal everywhere? Take for instance $f(x)=x^2$ and $g(x)=2x^2$.
$endgroup$
– Jean-Claude Arbaut
Dec 6 '18 at 1:22
$begingroup$
So how would I continue?
$endgroup$
– GustoCo
Dec 6 '18 at 1:45
$begingroup$
To show that $exp(x+y) = exp(x)exp(y)$, consider the function $f(x,y) = exp(x)exp(y)exp(-(x+y))$ and show that both first partials are $0$ everywhere, and $f(0,0)=1$, so the $f(x,y) equiv 1$, and so the equality holds. You can adapt this argument to your situation.
$endgroup$
– AlexanderJ93
Dec 6 '18 at 3:44
$begingroup$
To show that $exp(x+y) = exp(x)exp(y)$, consider the function $f(x,y) = exp(x)exp(y)exp(-(x+y))$ and show that both first partials are $0$ everywhere, and $f(0,0)=1$, so the $f(x,y) equiv 1$, and so the equality holds. You can adapt this argument to your situation.
$endgroup$
– AlexanderJ93
Dec 6 '18 at 3:44
add a comment |
$begingroup$
Given $f(x) = exp(ax)$, then the derivative, by definition is given by $frac{df}{dx} = aexp(ax)$. Also, it is known that $f(0) = 1$
I want to prove that $exp((a+b)x)$ is equal to $exp(ax)cdotexp(bx)$.
So, to do this, I take the derivative of $exp((a+b)x)$ which results in $(a+b)exp((a+b)x)$ by definition of the exponential. Next, I take the derivative of $exp(ax)cdotexp(bx)$ which results in $(a+b)exp(ax)cdotexp(bx)$.
Now, I evaluate $frac{d}{dx}$$[exp((a+b)x) - exp(ax)cdotexp(bx)]$ at $x = 0$. This results in zero.
My question now is, have I completed the proof. If yes, can someone explain why?
calculus exponential-function
edited Dec 6 '18 at 5:23
Tianlalu
3,08421138
asked Dec 6 '18 at 1:05
GustoCoGustoCo
428
$endgroup$
Given $f(x) = exp(ax)$, then the derivative, by definition is given by $frac{df}{dx} = aexp(ax)$. Also, it is known that $f(0) = 1$
I want to prove that $exp((a+b)x)$ is equal to $exp(ax)cdotexp(bx)$.
So, to do this, I take the derivative of $exp((a+b)x)$ which results in $(a+b)exp((a+b)x)$ by definition of the exponential. Next, I take the derivative of $exp(ax)cdotexp(bx)$ which results in $(a+b)exp(ax)cdotexp(bx)$.
Now, I evaluate $frac{d}{dx}$$[exp((a+b)x) - exp(ax)cdotexp(bx)]$ at $x = 0$. This results in zero.
My question now is, have I completed the proof. If yes, can someone explain why?
calculus exponential-function
calculus exponential-function
edited Dec 6 '18 at 5:23
Tianlalu
3,08421138
asked Dec 6 '18 at 1:05
GustoCoGustoCo
428
edited Dec 6 '18 at 5:23
Tianlalu
3,08421138
asked Dec 6 '18 at 1:05
GustoCoGustoCo
428
edited Dec 6 '18 at 5:23
Tianlalu
3,08421138
edited Dec 6 '18 at 5:23
Tianlalu
3,08421138
edited Dec 6 '18 at 5:23
Tianlalu
3,08421138
3,08421138
asked Dec 6 '18 at 1:05
GustoCoGustoCo
428
asked Dec 6 '18 at 1:05
GustoCoGustoCo
428
asked Dec 6 '18 at 1:05
GustoCoGustoCo
428
428
3
$begingroup$
No. You have two functions $f$ and $g$, that you differentiate, and you check the derivatives are equal at $0$. Does it mean $f$ and $g$ are equal everywhere? Take for instance $f(x)=x^2$ and $g(x)=2x^2$.
$endgroup$
– Jean-Claude Arbaut
Dec 6 '18 at 1:22
$begingroup$
So how would I continue?
$endgroup$
– GustoCo
Dec 6 '18 at 1:45
$begingroup$
To show that $exp(x+y) = exp(x)exp(y)$, consider the function $f(x,y) = exp(x)exp(y)exp(-(x+y))$ and show that both first partials are $0$ everywhere, and $f(0,0)=1$, so the $f(x,y) equiv 1$, and so the equality holds. You can adapt this argument to your situation.
$endgroup$
– AlexanderJ93
Dec 6 '18 at 3:44
$begingroup$
To show that $exp(x+y) = exp(x)exp(y)$, consider the function $f(x,y) = exp(x)exp(y)exp(-(x+y))$ and show that both first partials are $0$ everywhere, and $f(0,0)=1$, so the $f(x,y) equiv 1$, and so the equality holds. You can adapt this argument to your situation.
$endgroup$
– AlexanderJ93
Dec 6 '18 at 3:44
add a comment |
3
$begingroup$
No. You have two functions $f$ and $g$, that you differentiate, and you check the derivatives are equal at $0$. Does it mean $f$ and $g$ are equal everywhere? Take for instance $f(x)=x^2$ and $g(x)=2x^2$.
$endgroup$
– Jean-Claude Arbaut
Dec 6 '18 at 1:22
$begingroup$
So how would I continue?
$endgroup$
– GustoCo
Dec 6 '18 at 1:45
$begingroup$
To show that $exp(x+y) = exp(x)exp(y)$, consider the function $f(x,y) = exp(x)exp(y)exp(-(x+y))$ and show that both first partials are $0$ everywhere, and $f(0,0)=1$, so the $f(x,y) equiv 1$, and so the equality holds. You can adapt this argument to your situation.
$endgroup$
– AlexanderJ93
Dec 6 '18 at 3:44
$begingroup$
To show that $exp(x+y) = exp(x)exp(y)$, consider the function $f(x,y) = exp(x)exp(y)exp(-(x+y))$ and show that both first partials are $0$ everywhere, and $f(0,0)=1$, so the $f(x,y) equiv 1$, and so the equality holds. You can adapt this argument to your situation.
$endgroup$
– AlexanderJ93
Dec 6 '18 at 3:44
3
3
$begingroup$
No. You have two functions $f$ and $g$, that you differentiate, and you check the derivatives are equal at $0$. Does it mean $f$ and $g$ are equal everywhere? Take for instance $f(x)=x^2$ and $g(x)=2x^2$.
$endgroup$
– Jean-Claude Arbaut
Dec 6 '18 at 1:22
$begingroup$
No. You have two functions $f$ and $g$, that you differentiate, and you check the derivatives are equal at $0$. Does it mean $f$ and $g$ are equal everywhere? Take for instance $f(x)=x^2$ and $g(x)=2x^2$.
$endgroup$
– Jean-Claude Arbaut
Dec 6 '18 at 1:22
$begingroup$
So how would I continue?
$endgroup$
– GustoCo
Dec 6 '18 at 1:45
$begingroup$
So how would I continue?
$endgroup$
– GustoCo
Dec 6 '18 at 1:45
$begingroup$
To show that $exp(x+y) = exp(x)exp(y)$, consider the function $f(x,y) = exp(x)exp(y)exp(-(x+y))$ and show that both first partials are $0$ everywhere, and $f(0,0)=1$, so the $f(x,y) equiv 1$, and so the equality holds. You can adapt this argument to your situation.
$endgroup$
– AlexanderJ93
Dec 6 '18 at 3:44
$begingroup$
To show that $exp(x+y) = exp(x)exp(y)$, consider the function $f(x,y) = exp(x)exp(y)exp(-(x+y))$ and show that both first partials are $0$ everywhere, and $f(0,0)=1$, so the $f(x,y) equiv 1$, and so the equality holds. You can adapt this argument to your situation.
$endgroup$
– AlexanderJ93
Dec 6 '18 at 3:44
$begingroup$
To show that $exp(x+y) = exp(x)exp(y)$, consider the function $f(x,y) = exp(x)exp(y)exp(-(x+y))$ and show that both first partials are $0$ everywhere, and $f(0,0)=1$, so the $f(x,y) equiv 1$, and so the equality holds. You can adapt this argument to your situation.
$endgroup$
– AlexanderJ93
Dec 6 '18 at 3:44
$begingroup$
To show that $exp(x+y) = exp(x)exp(y)$, consider the function $f(x,y) = exp(x)exp(y)exp(-(x+y))$ and show that both first partials are $0$ everywhere, and $f(0,0)=1$, so the $f(x,y) equiv 1$, and so the equality holds. You can adapt this argument to your situation.
$endgroup$
– AlexanderJ93
Dec 6 '18 at 3:44
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Any function whose derivative is 0 for all $x$ is a constant function.
So what you have shown is that $e^{(a+b)x}-e^{ax}e^{bx}$ is a constant function. Any input, $x$, will yield the same output. So for example $e^{5(a+b)}-e^{5a}e^{5b}=e^{89(a+b)}-e^{89a}e^{88b}= e^{-2(a+b)}-e^{-2a}e^{-2b}$. It doesn’t matter what input we choose, it will give us the same thing.
So how do we find the function value? How do we use this information to show the result?
answered Dec 6 '18 at 5:41
John BJohn B
3087
$endgroup$
$begingroup$
Input the value and get the result.
$endgroup$
– William Elliot
Dec 6 '18 at 10:06
1
$begingroup$
Those questions were rhetorical. They were meant for the question asker to think about.
$endgroup$
– John B
Dec 6 '18 at 14:03
$begingroup$
I'm a little confused. I showed exp((a+b)x)−exp(ax)*exp(bx) is 0 when we sub in x = 0. The only reason I subbed in x = 0 is because the definition of the exponential stated at the beginning of the question allows for it. How do I extend this for all value of x? For example, from the definition, I only know that exp((a+b)0) - exp(a*0)*exp(b*0) = 1 - 1 = 0. How can I extend this to exp((a+b)*5) - exp(a*5)*exp(b*5) = 0?
$endgroup$
– GustoCo
Dec 6 '18 at 20:15
$begingroup$
Let me give you examples of constant functions. f(x)=1, g(x)=2, h(x)=-67. If I input 0 into these functions, I get f(0)=1, g(0)=2, h(0)=-67. If I input
$endgroup$
– John B
Dec 6 '18 at 20:24
$begingroup$
If I input 4 into these functions, I get f(4)=1, g(4)=2, h(4)=-67. If I input any number y into these functions, I get f(y)=1, g(y)=2, h(y)=-67. @GustoCo
$endgroup$
– John B
Dec 6 '18 at 20:27
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
Any function whose derivative is 0 for all $x$ is a constant function.
So what you have shown is that $e^{(a+b)x}-e^{ax}e^{bx}$ is a constant function. Any input, $x$, will yield the same output. So for example $e^{5(a+b)}-e^{5a}e^{5b}=e^{89(a+b)}-e^{89a}e^{88b}= e^{-2(a+b)}-e^{-2a}e^{-2b}$. It doesn’t matter what input we choose, it will give us the same thing.
So how do we find the function value? How do we use this information to show the result?
answered Dec 6 '18 at 5:41
John BJohn B
3087
$endgroup$
$begingroup$
Input the value and get the result.
$endgroup$
– William Elliot
Dec 6 '18 at 10:06
1
$begingroup$
Those questions were rhetorical. They were meant for the question asker to think about.
$endgroup$
– John B
Dec 6 '18 at 14:03
$begingroup$
I'm a little confused. I showed exp((a+b)x)−exp(ax)*exp(bx) is 0 when we sub in x = 0. The only reason I subbed in x = 0 is because the definition of the exponential stated at the beginning of the question allows for it. How do I extend this for all value of x? For example, from the definition, I only know that exp((a+b)0) - exp(a*0)*exp(b*0) = 1 - 1 = 0. How can I extend this to exp((a+b)*5) - exp(a*5)*exp(b*5) = 0?
$endgroup$
– GustoCo
Dec 6 '18 at 20:15
$begingroup$
Let me give you examples of constant functions. f(x)=1, g(x)=2, h(x)=-67. If I input 0 into these functions, I get f(0)=1, g(0)=2, h(0)=-67. If I input
$endgroup$
– John B
Dec 6 '18 at 20:24
$begingroup$
If I input 4 into these functions, I get f(4)=1, g(4)=2, h(4)=-67. If I input any number y into these functions, I get f(y)=1, g(y)=2, h(y)=-67. @GustoCo
$endgroup$
– John B
Dec 6 '18 at 20:27
add a comment |
$begingroup$
Any function whose derivative is 0 for all $x$ is a constant function.
So what you have shown is that $e^{(a+b)x}-e^{ax}e^{bx}$ is a constant function. Any input, $x$, will yield the same output. So for example $e^{5(a+b)}-e^{5a}e^{5b}=e^{89(a+b)}-e^{89a}e^{88b}= e^{-2(a+b)}-e^{-2a}e^{-2b}$. It doesn’t matter what input we choose, it will give us the same thing.
So how do we find the function value? How do we use this information to show the result?
answered Dec 6 '18 at 5:41
John BJohn B
3087
$endgroup$
$begingroup$
Input the value and get the result.
$endgroup$
– William Elliot
Dec 6 '18 at 10:06
1
$begingroup$
Those questions were rhetorical. They were meant for the question asker to think about.
$endgroup$
– John B
Dec 6 '18 at 14:03
$begingroup$
I'm a little confused. I showed exp((a+b)x)−exp(ax)*exp(bx) is 0 when we sub in x = 0. The only reason I subbed in x = 0 is because the definition of the exponential stated at the beginning of the question allows for it. How do I extend this for all value of x? For example, from the definition, I only know that exp((a+b)0) - exp(a*0)*exp(b*0) = 1 - 1 = 0. How can I extend this to exp((a+b)*5) - exp(a*5)*exp(b*5) = 0?
$endgroup$
– GustoCo
Dec 6 '18 at 20:15
$begingroup$
Let me give you examples of constant functions. f(x)=1, g(x)=2, h(x)=-67. If I input 0 into these functions, I get f(0)=1, g(0)=2, h(0)=-67. If I input
$endgroup$
– John B
Dec 6 '18 at 20:24
$begingroup$
If I input 4 into these functions, I get f(4)=1, g(4)=2, h(4)=-67. If I input any number y into these functions, I get f(y)=1, g(y)=2, h(y)=-67. @GustoCo
$endgroup$
– John B
Dec 6 '18 at 20:27
add a comment |
$begingroup$
Any function whose derivative is 0 for all $x$ is a constant function.
So what you have shown is that $e^{(a+b)x}-e^{ax}e^{bx}$ is a constant function. Any input, $x$, will yield the same output. So for example $e^{5(a+b)}-e^{5a}e^{5b}=e^{89(a+b)}-e^{89a}e^{88b}= e^{-2(a+b)}-e^{-2a}e^{-2b}$. It doesn’t matter what input we choose, it will give us the same thing.
So how do we find the function value? How do we use this information to show the result?
answered Dec 6 '18 at 5:41
John BJohn B
3087
$endgroup$
Any function whose derivative is 0 for all $x$ is a constant function.
So what you have shown is that $e^{(a+b)x}-e^{ax}e^{bx}$ is a constant function. Any input, $x$, will yield the same output. So for example $e^{5(a+b)}-e^{5a}e^{5b}=e^{89(a+b)}-e^{89a}e^{88b}= e^{-2(a+b)}-e^{-2a}e^{-2b}$. It doesn’t matter what input we choose, it will give us the same thing.
So how do we find the function value? How do we use this information to show the result?
answered Dec 6 '18 at 5:41
John BJohn B
3087
answered Dec 6 '18 at 5:41
John BJohn B
3087
answered Dec 6 '18 at 5:41
John BJohn B
3087
answered Dec 6 '18 at 5:41
John BJohn B
3087
3087
$begingroup$
Input the value and get the result.
$endgroup$
– William Elliot
Dec 6 '18 at 10:06
1
$begingroup$
Those questions were rhetorical. They were meant for the question asker to think about.
$endgroup$
– John B
Dec 6 '18 at 14:03
$begingroup$
I'm a little confused. I showed exp((a+b)x)−exp(ax)*exp(bx) is 0 when we sub in x = 0. The only reason I subbed in x = 0 is because the definition of the exponential stated at the beginning of the question allows for it. How do I extend this for all value of x? For example, from the definition, I only know that exp((a+b)0) - exp(a*0)*exp(b*0) = 1 - 1 = 0. How can I extend this to exp((a+b)*5) - exp(a*5)*exp(b*5) = 0?
$endgroup$
– GustoCo
Dec 6 '18 at 20:15
$begingroup$
Let me give you examples of constant functions. f(x)=1, g(x)=2, h(x)=-67. If I input 0 into these functions, I get f(0)=1, g(0)=2, h(0)=-67. If I input
$endgroup$
– John B
Dec 6 '18 at 20:24
$begingroup$
If I input 4 into these functions, I get f(4)=1, g(4)=2, h(4)=-67. If I input any number y into these functions, I get f(y)=1, g(y)=2, h(y)=-67. @GustoCo
$endgroup$
– John B
Dec 6 '18 at 20:27
add a comment |
$begingroup$
Input the value and get the result.
$endgroup$
– William Elliot
Dec 6 '18 at 10:06
1
$begingroup$
Those questions were rhetorical. They were meant for the question asker to think about.
$endgroup$
– John B
Dec 6 '18 at 14:03
$begingroup$
I'm a little confused. I showed exp((a+b)x)−exp(ax)*exp(bx) is 0 when we sub in x = 0. The only reason I subbed in x = 0 is because the definition of the exponential stated at the beginning of the question allows for it. How do I extend this for all value of x? For example, from the definition, I only know that exp((a+b)0) - exp(a*0)*exp(b*0) = 1 - 1 = 0. How can I extend this to exp((a+b)*5) - exp(a*5)*exp(b*5) = 0?
$endgroup$
– GustoCo
Dec 6 '18 at 20:15
$begingroup$
Let me give you examples of constant functions. f(x)=1, g(x)=2, h(x)=-67. If I input 0 into these functions, I get f(0)=1, g(0)=2, h(0)=-67. If I input
$endgroup$
– John B
Dec 6 '18 at 20:24
$begingroup$
If I input 4 into these functions, I get f(4)=1, g(4)=2, h(4)=-67. If I input any number y into these functions, I get f(y)=1, g(y)=2, h(y)=-67. @GustoCo
$endgroup$
– John B
Dec 6 '18 at 20:27
$begingroup$
Input the value and get the result.
$endgroup$
– William Elliot
Dec 6 '18 at 10:06
$begingroup$
Input the value and get the result.
$endgroup$
– William Elliot
Dec 6 '18 at 10:06
1
1
$begingroup$
Those questions were rhetorical. They were meant for the question asker to think about.
$endgroup$
– John B
Dec 6 '18 at 14:03
$begingroup$
Those questions were rhetorical. They were meant for the question asker to think about.
$endgroup$
– John B
Dec 6 '18 at 14:03
$begingroup$
I'm a little confused. I showed exp((a+b)x)−exp(ax)*exp(bx) is 0 when we sub in x = 0. The only reason I subbed in x = 0 is because the definition of the exponential stated at the beginning of the question allows for it. How do I extend this for all value of x? For example, from the definition, I only know that exp((a+b)0) - exp(a*0)*exp(b*0) = 1 - 1 = 0. How can I extend this to exp((a+b)*5) - exp(a*5)*exp(b*5) = 0?
$endgroup$
– GustoCo
Dec 6 '18 at 20:15
$begingroup$
I'm a little confused. I showed exp((a+b)x)−exp(ax)*exp(bx) is 0 when we sub in x = 0. The only reason I subbed in x = 0 is because the definition of the exponential stated at the beginning of the question allows for it. How do I extend this for all value of x? For example, from the definition, I only know that exp((a+b)0) - exp(a*0)*exp(b*0) = 1 - 1 = 0. How can I extend this to exp((a+b)*5) - exp(a*5)*exp(b*5) = 0?
$endgroup$
– GustoCo
Dec 6 '18 at 20:15
$begingroup$
Let me give you examples of constant functions. f(x)=1, g(x)=2, h(x)=-67. If I input 0 into these functions, I get f(0)=1, g(0)=2, h(0)=-67. If I input
$endgroup$
– John B
Dec 6 '18 at 20:24
$begingroup$
Let me give you examples of constant functions. f(x)=1, g(x)=2, h(x)=-67. If I input 0 into these functions, I get f(0)=1, g(0)=2, h(0)=-67. If I input
$endgroup$
– John B
Dec 6 '18 at 20:24
$begingroup$
If I input 4 into these functions, I get f(4)=1, g(4)=2, h(4)=-67. If I input any number y into these functions, I get f(y)=1, g(y)=2, h(y)=-67. @GustoCo
$endgroup$
– John B
Dec 6 '18 at 20:27
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If I input 4 into these functions, I get f(4)=1, g(4)=2, h(4)=-67. If I input any number y into these functions, I get f(y)=1, g(y)=2, h(y)=-67. @GustoCo
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– John B
Dec 6 '18 at 20:27
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3
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No. You have two functions $f$ and $g$, that you differentiate, and you check the derivatives are equal at $0$. Does it mean $f$ and $g$ are equal everywhere? Take for instance $f(x)=x^2$ and $g(x)=2x^2$.
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– Jean-Claude Arbaut
Dec 6 '18 at 1:22
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So how would I continue?
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– GustoCo
Dec 6 '18 at 1:45
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To show that $exp(x+y) = exp(x)exp(y)$, consider the function $f(x,y) = exp(x)exp(y)exp(-(x+y))$ and show that both first partials are $0$ everywhere, and $f(0,0)=1$, so the $f(x,y) equiv 1$, and so the equality holds. You can adapt this argument to your situation.
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– AlexanderJ93
Dec 6 '18 at 3:44
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To show that $exp(x+y) = exp(x)exp(y)$, consider the function $f(x,y) = exp(x)exp(y)exp(-(x+y))$ and show that both first partials are $0$ everywhere, and $f(0,0)=1$, so the $f(x,y) equiv 1$, and so the equality holds. You can adapt this argument to your situation.
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– AlexanderJ93
Dec 6 '18 at 3:44