Find the derivative of:

$$f(x) = \frac{e^{x^{2}} (\arcsin{x})^{2}x\sqrt{\cos{x}}}{(\ln{x})^{6} \sin^{2}x}$$

Attempted Solution

The most productive approach seems to be logarithmic differentiation:

$$\ln |f(x)| = \ln \left|\frac{e^{x^{2}} (\arcsin{x})^{2}x\sqrt{\cos{x}}}{(\ln{x})^{6} \sin^{2}x}\right|$$

Distributing the natural logarithm function gives addition instead of multiplication and subtraction instead of division:

$$\ln \left|\frac{e^{x^{2}} (\arcsin{x})^{2}x\sqrt{\cos{x}}}{(\ln{x})^{6} \sin^{2}x}\right| = \ln |e^{x^2}| + \ln |(\arcsin x)^2| + \ln |x|$$ $$+ \ln |\sqrt{\cos x}| - \ln |(\ln x)^6| - \ln |\sin^2 x|$$

Taking the derivative of both sides gives us:

$$f'(x) = f(x) \left( \frac{2 e^{x^2}}{e^{x^2}} + \frac{2 \arcsin x}{(\arcsin x)^2} \frac{1}{\sqrt{1-x^2}} + \frac{1}{x} + \frac{-\sin x}{2 \cos x} + \frac{6 (\ln x)^5}{(\ln x)^6} \frac{1}{x} - \frac{2 \sin x \cos x}{\sin x}\right) $$

Simplifying gives:

$$f'(x) = f(x) \left( 2 + \frac{2}{\arcsin x \sqrt{1-x^2}} + \frac{1}{x} -\frac{1}{2} \tan x + \frac{6}{x \ln x} + 2\cos x \right)$$

However, this is not the correct answer. In particular, the first and last terms are wrong. Where and how did it go wrong?

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    $\begingroup$ That should be $2x e^{x^2}$ rather than $2e^{x^2}$, and you wrote $\sin x$ instead of $\sin^2 x$ in the denominator of the last term. $\endgroup$ – user258700 Jul 28 '16 at 18:43
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    $\begingroup$ You don't need to take absolute values, the function is positive. $\endgroup$ – zhw. Jul 28 '16 at 18:47
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    $\begingroup$ You don't seem to be simplifying the $\ln$ before differentiating. Makes life harder, errors more likely. The first term is $\ln(e^{x^2})$, that is, $x^2$. Its derivative is $2x$. $\endgroup$ – André Nicolas Jul 28 '16 at 18:48

Take the natural log to get $$\ln f = \ln e^{x^2} + 2\ln \arcsin x + \ln x + \frac{1}{2} \ln \cos x - 6\ln \ln x - 2\ln \sin x$$

And then simplifying that a bit, i.e: $\ln e^{x^2} = x^2$ and differentiating gives $$f' = f\left(2x + \cdots - 2 \frac{\cos x}{\sin x}\right)$$

So that gives you the correct first and last terms. Remember that logs do more than just convert multiplication to addition, they turn powers to coefficients, making differentiation a lot easier.

  • $\begingroup$ Ah, cancelling the ln and e at once was a great point. I also seem to be laboring under the faulty delusion that e^(x^2) = e^(2x). Not really sure where I got that idea from. $\endgroup$ – MathInferno Jul 28 '16 at 18:51
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    $\begingroup$ It's actually a fairly common mistake from $(a^b)^c = a^{bc}$ so $(e^x)^2 = e^{2x}$ but of course here $e^{x^2} \neq (e^{x})^2$ $\endgroup$ – Zain Patel Jul 28 '16 at 18:53
  • $\begingroup$ ...because there is an implicit parenthesis $e^{x^2} = e^{(x^2)}$? $\endgroup$ – MathInferno Jul 28 '16 at 18:56
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    $\begingroup$ Yup! It's easy to see, for example that raising $2$ to $3^2 = 9$ which is $2^9 = 512$ is different from raising $2^3$ to $2$ which is $8^2 = 64$, there's a big difference in size between squaring and then putting in the power versus putting in the power and then squaring the whole thing. $\endgroup$ – Zain Patel Jul 28 '16 at 18:59

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