# Differentiate the following without using the number e

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

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

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

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

• your second needs a proof and its proof contains as i think e – Semsem Mar 3 '14 at 17:33
• 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. – 7raiden7 Mar 3 '14 at 17:38
• Now i got what you did – Semsem Mar 3 '14 at 17:42
• Hope it helps, I've always been fascinated by this form of differentiation, as you neglect all exponential forms! – 7raiden7 Mar 3 '14 at 17:45
• it appears that I cannot avoid logs. regardless, I gave all of you a thumbs up. – jax Mar 3 '14 at 18:15

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.

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., $$y'=y\left(\cos x\cdot \ln (2+\cos x)+\sin x \cdot \frac{2-\sin x}{2+\cos x}\right)$$ $$y'= (2+\cos x)^{\sin x}\left(\cos x\cdot \ln (2+\cos x)+\sin x \cdot \frac{2-\sin x}{2+\cos x}\right) .$$