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I am trying an FTC problem.

$$\frac{d}{dx}\int_0^{9\ln x}e^t \,dt=\ ?$$

How do I evaluate $$e^{9\ln(x)(9/x)}$$

EDIT: Ok, I made a mistake above, and put the chain rule result in the exponent!
Wrong! Here is the correct way, using FTC:

$$\huge{\frac{d}{dx}\int_0^{9\ln x}e^t \,dt=(e^{9lnx})\frac{9}{x}=(e^{lnx})^9(\frac{9}{x})=\frac{x^9}{1}\frac{9}{x}=9x^8}$$

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  • $\begingroup$ Try to avoid using \displaystyle in the title of the question, since it appears too big on the main page. $\endgroup$
    – user17762
    Dec 9, 2013 at 2:56

4 Answers 4

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$$ \huge{e^{9\ln(x)(\frac{9}{x})}=e^{\ln(x)\frac{81}{x}}=(e^{\ln(x)})^{\frac{81}{x}}=x^{\frac{81}{x}}}$$

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  • $\begingroup$ So, you claim that $x^{\frac{81}{x}}=x^2-9$? $\endgroup$ Dec 9, 2013 at 3:29
  • $\begingroup$ Thanks, rewriting power to a power was helpful in the solution. $\endgroup$
    – JackOfAll
    Dec 9, 2013 at 21:51
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I don't understand what you did, but anyway: $$\frac{d}{dx}\int_0^{9\ln x}e^t \,dt=\frac{d}{dx}(e^{9\ln x}-1)=\frac{d}{dx}(x^9-1)=9x^8.$$

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Hint: You can switch the order of the factors. $\exp(abc)=\exp(bac)$.

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Ok, I made a mistake above, and put the chain rule result in the exponent!
Wrong! Here is the correct way, using FTC:

$$\huge{\frac{d}{dx}\int_0^{9\ln x}e^t \,dt=(e^{9lnx})\frac{9}{x}=(e^{lnx})^9(\frac{9}{x})=\frac{x^9}{1}\frac{9}{x}=9x^8}$$

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