The derivative of $e^{\cos(x)}$ is $-\sin(x)e^{\cos(x)}$. However I would like to prove it using first principles, i.e. by using $f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$. I tried Taylor series but it didn't work out. Thanks for any help.
[Math] Calculation of the derivative of $e^{\cos(x)}$ from first principles
calculusderivatives
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Best Answer
When we have a question of calculating the derivative via first principles then it means that the idea is to drill down the definition of derivative via actual examples. It also signifies that the student is beginning to learn differential calculus. It is therefore much better to use techniques which rely on standard limits and don't rely on advanced theorems of differential calculus (like Taylor's series on L'Hospital's rule).
We will make use of the following standard limits $$\lim_{h \to 0}\frac{e^{h} - 1}{h} = 1, \,\,\lim_{h \to 0}\frac{\sin h}{h} = 1$$ We have $f(x) = e^{\cos x}$ and by definition of derivative $$\begin{aligned}f'(x)\, &= \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}\\ &= \lim_{h \to 0}\frac{e^{\cos(x + h)} - e^{\cos x}}{h}\\ &= \lim_{h \to 0}\frac{e^{\cos x}\{e^{\cos(x + h) - \cos x} - 1\}}{h}\\ &= e^{\cos x}\lim_{h \to 0}\frac{e^{\cos(x + h) - \cos x} - 1}{\cos(x + h) - \cos x}\cdot\frac{\cos(x + h) - \cos x}{h}\\ &= e^{\cos x}\lim_{t \to 0}\frac{e^{t} - 1}{t}\cdot\lim_{h \to 0}\frac{\cos(x + h) - \cos x}{h}\\ &= e^{\cos x}\cdot 1\cdot\lim_{h \to 0}\dfrac{-2\sin\left(x + \dfrac{h}{2}\right)\sin\left(\dfrac{h}{2}\right)}{h}\\ &= -e^{\cos x}\cdot\lim_{h \to 0}\sin\left(x + \frac{h}{2}\right)\cdot\lim_{h \to 0}\frac{\sin(h/2)}{h/2}\\ &= -e^{\cos x}\cdot\sin x\cdot 1 = -e^{\cos x}\sin x\end{aligned}$$ In the above derivation we have made the substitution $t = \cos(x + h) - \cos x$ and $t \to 0$ as $h \to 0$.