How to get derivative of integral with $ln(x)$?












0












$begingroup$


Find $f'(x)$ if $f(x) = int_1^{ln(x)} e^{t^2} ,dt$



The correct way to solve it:



$$f'(x) = e^{(ln{x})^2} frac{1}{x}$$



$$f'(x) = frac{1}{x}e^{(ln{x})^2}$$



I haven't seen an example like this before, but it seems relatively easy to solve.



I'm assuming the exponent $t$ gets replaced with $ln x$ and the $frac{1}{x}$ comes from the derivative of $ln x$ via the chain rule? But what are the exact steps to take to get the derivative of this integral?










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  • 3




    $begingroup$
    That's correct, and this is just a one-line application of the fundamental theorem of calculus. If $F$ is the antiderivative of $f$, you're just being asked to differentiate the function $x mapsto F(ln x)$.
    $endgroup$
    – T. Bongers
    Dec 11 '18 at 0:19
















0












$begingroup$


Find $f'(x)$ if $f(x) = int_1^{ln(x)} e^{t^2} ,dt$



The correct way to solve it:



$$f'(x) = e^{(ln{x})^2} frac{1}{x}$$



$$f'(x) = frac{1}{x}e^{(ln{x})^2}$$



I haven't seen an example like this before, but it seems relatively easy to solve.



I'm assuming the exponent $t$ gets replaced with $ln x$ and the $frac{1}{x}$ comes from the derivative of $ln x$ via the chain rule? But what are the exact steps to take to get the derivative of this integral?










share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    That's correct, and this is just a one-line application of the fundamental theorem of calculus. If $F$ is the antiderivative of $f$, you're just being asked to differentiate the function $x mapsto F(ln x)$.
    $endgroup$
    – T. Bongers
    Dec 11 '18 at 0:19














0












0








0





$begingroup$


Find $f'(x)$ if $f(x) = int_1^{ln(x)} e^{t^2} ,dt$



The correct way to solve it:



$$f'(x) = e^{(ln{x})^2} frac{1}{x}$$



$$f'(x) = frac{1}{x}e^{(ln{x})^2}$$



I haven't seen an example like this before, but it seems relatively easy to solve.



I'm assuming the exponent $t$ gets replaced with $ln x$ and the $frac{1}{x}$ comes from the derivative of $ln x$ via the chain rule? But what are the exact steps to take to get the derivative of this integral?










share|cite|improve this question









$endgroup$




Find $f'(x)$ if $f(x) = int_1^{ln(x)} e^{t^2} ,dt$



The correct way to solve it:



$$f'(x) = e^{(ln{x})^2} frac{1}{x}$$



$$f'(x) = frac{1}{x}e^{(ln{x})^2}$$



I haven't seen an example like this before, but it seems relatively easy to solve.



I'm assuming the exponent $t$ gets replaced with $ln x$ and the $frac{1}{x}$ comes from the derivative of $ln x$ via the chain rule? But what are the exact steps to take to get the derivative of this integral?







calculus integration derivatives definite-integrals






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share|cite|improve this question











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asked Dec 11 '18 at 0:15









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  • 3




    $begingroup$
    That's correct, and this is just a one-line application of the fundamental theorem of calculus. If $F$ is the antiderivative of $f$, you're just being asked to differentiate the function $x mapsto F(ln x)$.
    $endgroup$
    – T. Bongers
    Dec 11 '18 at 0:19














  • 3




    $begingroup$
    That's correct, and this is just a one-line application of the fundamental theorem of calculus. If $F$ is the antiderivative of $f$, you're just being asked to differentiate the function $x mapsto F(ln x)$.
    $endgroup$
    – T. Bongers
    Dec 11 '18 at 0:19








3




3




$begingroup$
That's correct, and this is just a one-line application of the fundamental theorem of calculus. If $F$ is the antiderivative of $f$, you're just being asked to differentiate the function $x mapsto F(ln x)$.
$endgroup$
– T. Bongers
Dec 11 '18 at 0:19




$begingroup$
That's correct, and this is just a one-line application of the fundamental theorem of calculus. If $F$ is the antiderivative of $f$, you're just being asked to differentiate the function $x mapsto F(ln x)$.
$endgroup$
– T. Bongers
Dec 11 '18 at 0:19










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