Question

question 2
find the derivative of $g(x)=\frac{x^{2}}{cos(x)}$.
$g(x)=-\frac{2x}{sin(x)}$
$g(x)=\frac{2xcos(x)+x^{2}sin(x)}{cos^{2}(x)}$
$g(x)=\frac{x - sin(x)}{cos(x)}$
$g(x)=-\frac{x^{2}sin(x)}{cos^{2}(x)}$

Explanation:

Step1: Recall quotient - rule

The quotient - rule states that if $g(x)=\frac{u(x)}{v(x)}$, then $g^{\prime}(x)=\frac{u^{\prime}(x)v(x)-u(x)v^{\prime}(x)}{v^{2}(x)}$. Here, $u(x) = x^{2}$ and $v(x)=\cos(x)$.

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

The derivative of $u(x)=x^{2}$ using the power - rule ($(x^{n})^\prime=nx^{n - 1}$) is $u^{\prime}(x) = 2x$. The derivative of $v(x)=\cos(x)$ is $v^{\prime}(x)=-\sin(x)$.

Step3: Apply the quotient - rule

Substitute $u(x),u^{\prime}(x),v(x),v^{\prime}(x)$ into the quotient - rule formula:
\[

$$\begin{align*} g^{\prime}(x)&=\frac{2x\cdot\cos(x)-x^{2}\cdot(-\sin(x))}{\cos^{2}(x)}\\ &=\frac{2x\cos(x)+x^{2}\sin(x)}{\cos^{2}(x)} \end{align*}$$

\]

Answer:

$g^{\prime}(x)=\frac{2x\cos(x)+x^{2}\sin(x)}{\cos^{2}(x)}$ (the second option)