\[\begin{aligned} \frac{d}{dx}(x^{\alpha}) &= \alpha x^{\alpha-1} \\ \frac{d}{dx} ( e^x ) &= e^x \end{aligned}\]
Example. \[\frac{d}{dx} (\sqrt{3x} + e^{x+2} ) = \frac{d}{dx} (\sqrt{3} x^{1/2} + e^2 e^x) = \sqrt{3} \frac{d}{dx} (x^{1/2}) + e^2 \frac{d}{dx} (e^x) = \frac{\sqrt{3}}{2}x^{-1/2}+e^{2+x}\] (Remember that \(e^{x+2} = e^x e^2\))
\[\frac{d}{dx}[f(x)g(x)] = g(x) \frac{d}{dx}f(x) + f(x) \frac{d}{dx} g(x)\] or in prime notation \[(fg)' = gf' + fg'\]
Example 1.
If \(f(x)=xe^x\), find \(f'(x).\)
Sol:
\[f'(x)= \frac{d}{dx}(x) e^x + x \frac{d}{dx}(e^x) = 1\cdot e^x + x e^x = (1+x)e^x\]
Example 2.
If \(f(x)=\sqrt{x}(1+x)\), find \(f'(x).\)
Sol:
\[f'(x)= (1+x)\frac{d}{dx}(\sqrt{x}) + \sqrt{x} \frac{d}{dx}(1+x) = (1+x)\cdot (\frac{1}{2}x^{-1/2}) + \sqrt{x}\cdot 1 = \frac{1}{2\sqrt{x}}+\frac{3}{2}\sqrt{x}\]
Shortcut: note that \(f(x)=\sqrt{x}+x^{3/2}\)
and compute \(f'\) using power rule.
Example 3.
If \(f(x) = e^x g(x)\), where \(g(0)=2\) and \(g'(0)=-10\).
Find \(f'(0)\).
Sol:
\[f'(x) = g(x) \frac{d}{dx}(e^x) + e^x \frac{d}{dx} g(x)
= e^x g(x)+ e^x g'(x). \]
Therefore, \(f'(0) = e^0g(0)+e^0g'(0)=2+(-10)=-8.\)
\[\dfrac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \dfrac{g(x)\frac{d}{dx}f(x) - f(x)\frac{d}{dx}g(x)}{[g(x)]^2}\] or in prime notation \[\left(\frac{f}{g}\right) = \dfrac{gf' - fg'}{g^2}\]
Example 1.
Differentiate \(H(x) = \frac{x^2-1}{x^3+1}.\)
Sol: \(H = \frac{f}{g}\), where
\(f(x)=x^2-1\), \(f'(x)=2x\),
\(g(x)=x^3+1\), \(g'(x)=3x^2\).
We deduce that
\[H'(x) = \dfrac{(x^3+1)(2x)-(x^2-1)(3x^2)}{(x^3+1)^2}\]
Example 2.
Find the equation of the tangent line to the curve \(H(x)=\frac{x}{e^x+1}\) at the point \((0,0)\) ?
Sol: The tangent line at \((0,0)\) is:
\[y=k(x-0)+0=kx,\]
where the slope \(k = H'(0)\).
It suffices to find the value of \(H'(0)\).
\(H=\frac{f}{g}\),
where \(f(x)=x\), \(f'(x)=1\), \(g(x)=e^x+1\),
\(g'(x)=e^x\).
Then we have
\[H'(x)=\frac{(e^x+1)-xe^x}{(e^x+1)^2}.\]
Hence the slope is \(k=H'(0) = \frac{(1+1)-0}{(1+1)^2} = \frac{1}{2}\).
The equation of the tangent line at \((0,0)\) is:
\[y=\frac{1}{2}x.\]
Ex. Use quotient rule to show that \[\dfrac{1}{g(x)} = -\dfrac{g'(x)}{[g(x)]^2}\]