Applying Differentiation Rules to Exponential Functions
We have learned that the derivative of an exponential function is given by the following formulas:
\[\begin{array} &y=e^x &\implies y'=e^x\\ y=e^{f(x)} &\implies y'=e^{f(x)}f'(x)\\ y=a^x &\implies y'=a^x(\ln a)\\ y=a^{f(x)} &\implies y'=a^{f(x)}(\ln a)\cdot f'(x), \end{array}\]
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:
\[\begin{array} &y=cf(x) &\implies y'=cf'(x)\\ y=f(x)+g(x) &\implies y'=f'(x)+g'(x)\\ y=f(x)g(x) &\implies y'=f'(x)g(x)+f(x)g'(x)\\ y=\frac{f(x)}{g(x)} &\implies y'=\frac{g(x)f'(x)-f(x)g'(x)}{\big(g(x)\big)^2}\\ y=g\big(f(x)\big) &\implies y'=g'\big(f(x)\big)\cdot f'(x). \end{array}\]
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 \(y=3^{x^3+1}?\)
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
\[\begin{align} y'&=3^{x^3+1} (\ln 3) \cdot \left(x^3+1\right)'\\ &=3^{x^3+1} (\ln 3) \cdot 3x^2\\ &=3^{x^3+2}x^2 \cdot \ln 3. \ _\square \end{align}\]
What is the derivative of \(y=5^{\sin x}?\)
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
\[\begin{align} y'&=5^{\sin x}(\ln 5) \cdot (\sin x)'\\ &=5^{\sin x}(\ln 5) \cdot (\cos x)\\ &=5^{\sin x} (\cos x) \cdot (\ln 5). \ _\square \end{align}\]
What is the derivative of \(f(x)=5^{\cos3x }?\)
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
\[\begin{align} f(x)&=5^{\cos3x }\\&=e^{\ln5^{\cos3x}}\\&=e^{(\cos3x)\cdot(\ln5)}\\\\ \Rightarrow f'(x)&=-3\sin3x\cdot\ln5\cdot e^{(\cos3x)\cdot(\ln5)}.\ _\square \end{align}\]
What is the derivative of \(y=xe^{\cos x}?\)
We have
\[\begin{align} y'&=(x)'e^{\cos x}+x\left(e^{\cos x}\right)'\\ &=1\cdot e^{\cos x}+x \cdot e^{\cos x}\cdot(\cos x)'\\ &= e^{\cos x}+x\cdot e^{\cos x}\cdot(-\sin x)\\ &= e^{\cos x}(1-x\sin x).\ _\square \end{align}\]
What is the derivative of \(y=e^{3 x}\sin 2x?\)
We have
\[\begin{align} y'&=\left(e^{3 x}\right)'\sin 2x+e^{3 x}(\sin 2x)'\\ &=3e^{3 x}\sin 2x+e^{3 x}\cdot 2\cos 2x\\ &=e^{3x}(3\sin 2x+2\cos 2x).\ _\square \end{align}\]
What is the derivative of \(\displaystyle{y=\frac{e^x-e^{-x}}{e^x+e^{-x}}?}\)
We have
\[\begin{align} y'&=\frac{\left(e^x-e^{-x}\right)'\left(e^x+e^{-x}\right)-\left(e^x-e^{-x}\right)\left(e^x+e^{-x}\right)'}{\left(e^x+e^{-x}\right)^2}\\\\ &=\frac{\left(e^x+e^{-x}\right)\left(e^x+e^{-x}\right)-\left(e^x-e^{-x}\right)\left(e^x-e^{-x}\right)}{\left(e^x+e^{-x}\right)^2}\\\\ &=\frac{\left(e^x+e^{-x}\right)^2-\left(e^x-e^{-x}\right)^2}{\left(e^x+e^{-x}\right)^2}\\\\ &=\frac{4}{\left(e^x+e^{-x}\right)^2}.\ _\square \end{align}\]