Applying Differentiation Rules to Exponential Functions

We have learned that the derivative of an exponential function is given by the following formulas:

$$\begin{array}{c}&y=e^x &\implies y'=e^x\\ y=e^{f(x)} &\implies y'=e^{f(x)}f'(x)\\ y=a^x &\implies y'=a^x(\ln a)\\ y=a^{f(x)} &\implies y'=a^{f(x)}(\ln a)\cdot f'(x), \end{array}$$

where we denote $\frac{dy}{dx}$ as $y'$ and $\frac{df(x)}{dx}$ as $f'(x),$ and $a$ is a random base such that $a>0$ and $a\ne1.$ When differentiating complex exponential functions, just stick to the formulas above along with the differentiation rules that we have learned earlier. To remind, here is a list of some differentiation rules:

$$\begin{array}{c}&y=cf(x) &\implies y'=cf'(x)\\ y=f(x)+g(x) &\implies y'=f'(x)+g'(x)\\ y=f(x)g(x) &\implies y'=f'(x)g(x)+f(x)g'(x)\\ y=\frac{f(x)}{g(x)} &\implies y'=\frac{g(x)f'(x)-f(x)g'(x)}{\big(g(x)\big)^2}\\ y=g\big(f(x)\big) &\implies y'=g'\big(f(x)\big)\cdot f'(x). \end{array}$$

Another rule that might be handy is one that we have learned when studying logarithms in algebra. Recall that

$$a^{\log_b c}=c^{\log_b a}.$$

Now let's try differentiating complex exponential functions using these rules.

What is the derivative of $y=3^{x^3+1}?$

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If $y=a^{f(x)},$ then $y'=a^{f(x)}(\ln a)\cdot f'(x).$ Since $y=3^{x^3+1}$ in this problem, we have

$$\begin{aligned} y'&=3^{x^3+1} (\ln 3) \cdot \left(x^3+1\right)'\\ &=3^{x^3+1} (\ln 3) \cdot 3x^2\\ &=3^{x^3+2}x^2 \cdot \ln 3. \ _\square \end{aligned}$$

What is the derivative of $y=5^{\sin x}?$

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If $y=a^{f(x)},$ then $y'=a^{f(x)}(\ln a)\cdot f'(x).$ Since $y=5^{\sin x}$ in this problem, we have

$$\begin{aligned} y'&=5^{\sin x}(\ln 5) \cdot (\sin x)'\\ &=5^{\sin x}(\ln 5) \cdot (\cos x)\\ &=5^{\sin x} (\cos x) \cdot (\ln 5). \ _\square \end{aligned}$$

What is the derivative of $f(x)=5^{\cos3x }?$

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This problem can be solved using the same method as example #2. However, here is another way to look at it.

Converting the base to $e$ by using the formula $e^{\ln a^x} =a^x$ gives

$$\begin{aligned} f(x)&=5^{\cos3x }\\&=e^{\ln5^{\cos3x}}\\&=e^{(\cos3x)\cdot(\ln5)}\\\\ \Rightarrow f'(x)&=-3\sin3x\cdot\ln5\cdot e^{(\cos3x)\cdot(\ln5)}.\ _\square \end{aligned}$$

What is the derivative of $y=xe^{\cos x}?$

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We have

$$\begin{aligned} y'&=(x)'e^{\cos x}+x\left(e^{\cos x}\right)'\\ &=1\cdot e^{\cos x}+x \cdot e^{\cos x}\cdot(\cos x)'\\ &= e^{\cos x}+x\cdot e^{\cos x}\cdot(-\sin x)\\ &= e^{\cos x}(1-x\sin x).\ _\square \end{aligned}$$

What is the derivative of $y=e^{3 x}\sin 2x?$

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We have

$$\begin{aligned} y'&=\left(e^{3 x}\right)'\sin 2x+e^{3 x}(\sin 2x)'\\ &=3e^{3 x}\sin 2x+e^{3 x}\cdot 2\cos 2x\\ &=e^{3x}(3\sin 2x+2\cos 2x).\ _\square \end{aligned}$$

What is the derivative of $\displaystyle{y=\frac{e^x-e^{-x}}{e^x+e^{-x}}?}$

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We have

$$\begin{aligned} y'&=\frac{\left(e^x-e^{-x}\right)'\left(e^x+e^{-x}\right)-\left(e^x-e^{-x}\right)\left(e^x+e^{-x}\right)'}{\left(e^x+e^{-x}\right)^2}\\\\ &=\frac{\left(e^x+e^{-x}\right)\left(e^x+e^{-x}\right)-\left(e^x-e^{-x}\right)\left(e^x-e^{-x}\right)}{\left(e^x+e^{-x}\right)^2}\\\\ &=\frac{\left(e^x+e^{-x}\right)^2-\left(e^x-e^{-x}\right)^2}{\left(e^x+e^{-x}\right)^2}\\\\ &=\frac{4}{\left(e^x+e^{-x}\right)^2}.\ _\square \end{aligned}$$