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I know how differentiate this using $e$ however, I was wondering if someone can differentiate it without using $e$.

$$\frac{d}{dx}(2+\cos x)^{\sin x}$$

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    $\begingroup$ Did you mean without logarithm, too? $\endgroup$ – lab bhattacharjee Mar 3 '14 at 17:06
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    $\begingroup$ Set $u=2+\cos x$, $v=\sin x$, and $f(u,v)=u^v$. Then, $${d\over dx} f(u,v)={\partial f\over \partial u}{du\over dx}+{\partial f\over \partial v}{dv\over dx}.$$ $\endgroup$ – David Mitra Mar 3 '14 at 17:07
  • $\begingroup$ Alternatively, use ${d\over dx} \ln f(x)={f'(x)\over f(x)}$. (I don't think you can avoid logs.) $\endgroup$ – David Mitra Mar 3 '14 at 17:22
  • $\begingroup$ I think your first claim is not true since u and v are dependent variables@DavidMitra $\endgroup$ – Semsem Mar 3 '14 at 17:31
  • $\begingroup$ @Semsem It's the multivariable chain rule. See case 1 here. $\endgroup$ – David Mitra Mar 3 '14 at 17:46
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In general: $$ \frac{d}{dx}f(x)^{g(x)}=\frac{d}{dx}\left(e^{g\ln f}\right)= $$ $$ =f(x)^{g(x)}\left(\cdot\frac{g(x)f'(x)}{f(x)}+ g'(x)\ln(f(x))\right). $$

In your case $f(x)=\cos(x)+2$, $g(x)=\sin(x)$, thus: $$ \frac{d}{dx}f(x)^{g(x)}=(2+\cos x)^{\sin x}\cdot\left(\frac{-\sin^2 x}{\cos(x)+2}+(\cos x\ln(2+\cos(x)))\right). $$

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  • $\begingroup$ your second needs a proof and its proof contains as i think e $\endgroup$ – Semsem Mar 3 '14 at 17:33
  • $\begingroup$ It does not need a proof, as it is based on chain rule and on elementary logarithm properties. Moreover, as long as you study exponential functions, $e$ is always concerned. $\endgroup$ – 7raiden7 Mar 3 '14 at 17:38
  • $\begingroup$ Now i got what you did $\endgroup$ – Semsem Mar 3 '14 at 17:42
  • $\begingroup$ Hope it helps, I've always been fascinated by this form of differentiation, as you neglect all exponential forms! $\endgroup$ – 7raiden7 Mar 3 '14 at 17:45
  • $\begingroup$ it appears that I cannot avoid logs. regardless, I gave all of you a thumbs up. $\endgroup$ – jax Mar 3 '14 at 18:15
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You are probably already familiar with the following four formulas:

$(a^x)'=a^x\cdot\ln a\quad=>\quad\ \ \Big[a^{f(x)}\Big]'=\ \ a^{f(x)}\cdot\ln a\cdot f'(x)$

$(x^n)'=n\cdot x^{n-1}\quad=>\quad\Big[g^n(x)\Big]'=n\cdot g^{n-1}(x)\cdot g'(x)$

Now, what you have to do is the following:

  • Pretend that the base were a constant, instead of a function in x, and derive the expression according to the first formula above.
  • Then pretend that the exponent were a constant, instead of function in x, and derive the expression according to the second formula above.
  • Then add these two derivative expressions together, in order to get the final answer.
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If $ y=(2+\cos x)^{\sin x}$, then $\ln y = \ln((2+\cos x)^{\sin x})=\sin x\cdot \ln (2+\cos x)$.

Then, using implicit derivation, $$\frac{y'}{y}=\cos x\cdot \ln(2+\cos x)+\sin x\cdot \frac{2-\sin x }{2+\cos x}, $$ i.e., $$y'=y\left(\cos x\cdot \ln (2+\cos x)+\sin x \cdot \frac{2-\sin x}{2+\cos x}\right) $$ $$y'= (2+\cos x)^{\sin x}\left(\cos x\cdot \ln (2+\cos x)+\sin x \cdot \frac{2-\sin x}{2+\cos x}\right) . $$

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