Question

1. (2 points) find the derivative of each function. pick 2.

a. $f(x)=x^{2}\tan x$
b. $g(x)=sin^{7}(3x)$
c. $h(x)=arcsin(2x + 1)$
d. $p(x)=arctan(3x + 2)$

Explanation:

Step1: Use product - rule for \(f(x)\)

The product - rule states that if \(y = u\cdot v\), then \(y^\prime=u^\prime v + uv^\prime\). For \(f(x)=x^{2}\tan x\), let \(u = x^{2}\) and \(v=\tan x\). We know that \(u^\prime = 2x\) and \(v^\prime=\sec^{2}x\). Then \(f^\prime(x)=(x^{2})^\prime\tan x+x^{2}(\tan x)^\prime=2x\tan x + x^{2}\sec^{2}x\).

Step2: Use chain - rule for \(g(x)\)

The chain - rule states that if \(y = f(u)\) and \(u = g(x)\), then \(y^\prime=f^\prime(u)\cdot g^\prime(x)\). For \(g(x)=\sin^{7}(3x)\), let \(u = \sin(3x)\), so \(g(x)=u^{7}\). First, \(\frac{dg}{du}=7u^{6}\), and for \(u=\sin(3x)\), \(\frac{du}{dx}=3\cos(3x)\). Then \(g^\prime(x)=7\sin^{6}(3x)\cdot3\cos(3x)=21\sin^{6}(3x)\cos(3x)\).

Answer:

For \(f(x)=x^{2}\tan x\), \(f^\prime(x)=2x\tan x + x^{2}\sec^{2}x\)
For \(g(x)=\sin^{7}(3x)\), \(g^\prime(x)=21\sin^{6}(3x)\cos(3x)\)