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Section Differentiation
Throughout the DEs course, we will be differentiating AND integrating. Letβs review a few common rules used for differentiation.
Product Rule: \(\ds\qquad \frac{d}{dx}[f(x) \cdot g(x)] = f(x)\cdot \frac{d}{dx}g(x) + g(x)\cdot \frac{d}{dx}f(x)\)
\(\displaystyle \)
Quotient Rule: \(\ds\qquad \frac{d}{dx}\left[\ds \frac{f(x)}{g(x)}\right] = \ds \frac{g(x) \cdot \ds \frac{d}{dx}f(x) - f(x) \cdot \ds \frac{d}{dx}g(x)}{[g(x)]^2}\)
\(\displaystyle \)
Chain Rule: \(\ds\qquad \frac{d}{dx}f\big(g(x)\big) = f'\big(g(x)\big) \cdot g'(x)\)
Consider the following example.
π Example 321 . Find the derivative of \(h(x) = (x^5 + 4)\sin(e^x)\) .
We have a product of functions, so we will need to use the product rule. We also have a composition of functions because \(e^x\) is inside of an sine function, so we will need to use the chain rule. Hereβs how it works.
\begin{align*}
h'(x) \amp = \ub{\frac{d}{dx} \Big[\ub{(x^5 + 4)}_{f} \cdot
\ub{\sin(e^x)}_{g} \Big]}_{
\knowl{./knowl/xref/product_rule.html}{\text{product rule}}} \\
\amp = (x^5+4)\cdot
\knowl{./knowl/xref/chain_rule.html}{\text{\(\ds\ub{\frac{d}{dx}\Big( \sin(e^x) \Big)}_{\text{chain rule}}\)}}
+ \sin(e^x) \cdot \frac{d}{dx}\Big( x^5 + 4 \Big) \\
\amp = (x^5+4)\cdot\cos(e^x) \cdot \frac{d}{dx}\Big( e^x \Big)
+ \sin(e^x) \cdot 5x^4 \\
\amp = (x^5+4)\cdot\cos(e^x) \cdot e^x + 5x^4\sin(e^x) \\
\amp = (x^5+4)e^x\cos(e^x) + 5x^4\sin(e^x)
\end{align*}
Find the indicated derivatives.
\(\ds \frac{d}{dz}\Big( e^{3z+6} \Big)\)
Solution
\begin{align*}
\frac{d}{dz}\Big( e^{3z+6} \Big)
\amp = e^{3z+6} \cdot \frac{d}{dz}\Big( 3z+6 \Big) \qquad
\knowl{./knowl/xref/chain_rule.html}{\text{chain rule}} \\
\amp = e^{3z+6} \cdot (3) \\
\amp = 3e^{3z+6}
\end{align*}
\begin{equation*}
3e^{3z+6}
\end{equation*}
\(\ds \frac{d}{dx}\Big( x^2e^{(\sin x)} \Big) \qquad\)
\begin{align*}
\knowl{./knowl/xref/product_rule.html}{\text{\(\ds\frac{d}{dx}\Big( x^2e^{(\sin x)} \Big)\)}}
\amp = x^2 \cdot
\knowl{./knowl/xref/chain_rule.html}{\text{\(\ds\frac{d}{dx}\Big(e^{(\sin x)} \Big)\)}}
+ e^{(\sin x)}\cdot \frac{d}{dx}\Big(x^2\Big) \\
\amp = x^2 \cdot e^{(\sin x)}\cdot\frac{d}{dx}\Big(\sin x \Big)
+ e^{(\sin x)}\cdot 2x \\
\amp = x^2\cdot e^{(\sin x)}\cdot\cos x + 2xe^{(\sin x)} \\
\amp = x^2 e^{(\sin x)}\cos x + 2xe^{(\sin x)}
\end{align*}
\begin{equation*}
x^2 e^{(\sin x)}\cos x + 2xe^{(\sin x)}
\end{equation*}
\(\ds \frac{d}{dt}\Big( (t+t^2)^{-1} \Big) \qquad\)
\begin{align*}
\frac{d}{dt}\Big( (t+t^2)^{-1} \Big)
\amp = (-1)(t+t^2)^{-2} \cdot \frac{d}{dt}\Big( t+t^2 \Big) \qquad
\knowl{./knowl/xref/chain_rule.html}{\text{chain rule}} \\
\amp = (-1)(t+t^2)^{-2} \cdot (1 + 2t) \\
\amp = \frac{-(1+2t)}{(t+t^2)^2}
\end{align*}
\begin{equation*}
\frac{-(1+2t)}{(t+t^2)^2}
\end{equation*}
\(\ds \frac{d}{ds}\Big( \ln(4s+7) \Big) \qquad\)
\begin{align*}
\frac{d}{ds}\Big( \ln(4s+7) \Big)
\amp = \frac{1}{4s+7} \cdot \frac{d}{ds}\Big( 4s+7 \Big) \qquad
\knowl{./knowl/xref/chain_rule.html}{\text{chain rule}} \\
\amp = \frac{1}{4s+7} \cdot (4) \\
\amp = \frac{4}{4s+7}
\end{align*}
\begin{equation*}
\frac{4}{4s+7}
\end{equation*}
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