Question

d. use quotient rule to differentiate $g(x)=\frac{e^{2 x}}{\left(x^{2}+1\
ight)^{10}}$

Explanation:

Step1: Recall the quotient rule

The quotient rule states that if \(g(x)=\frac{u(x)}{v(x)}\), then \(g^{\prime}(x)=\frac{u^{\prime}(x)v(x)-u(x)v^{\prime}(x)}{v(x)^{2}}\). Here, \(u(x) = e^{2x}\), \(v(x)=(x^{2}+1)^{10}\).

Step2: Find \(u^{\prime}(x)\)

Using the chain - rule, if \(u(x)=e^{2x}\), then \(u^{\prime}(x)=e^{2x}\cdot\frac{d}{dx}(2x)=2e^{2x}\).

Step3: Find \(v^{\prime}(x)\)

Using the chain - rule, if \(v(x)=(x^{2}+1)^{10}\), let \(t = x^{2}+1\), then \(v(x)=t^{10}\). \(\frac{dv}{dt}=10t^{9}\) and \(\frac{dt}{dx}=2x\). So \(v^{\prime}(x)=10(x^{2}+1)^{9}\cdot2x = 20x(x^{2}+1)^{9}\).

Step4: Apply the quotient rule

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Answer:

\(g^{\prime}(x)=\frac{e^{2x}(2x^{2}-20x + 2)}{(x^{2}+1)^{11}}\)