hw13 derivatives of trigonometric functions ($3.5)\nscore: 4/7 answered: 4/7\nquestion 5\nif…

Explanation:

Step1: Simplify the function

First, recall that $\tan x=\frac{\sin x}{\cos x}$ and $\sec x = \frac{1}{\cos x}$. So $f(x)=\frac{5x^{2}\tan x}{\sec x}=5x^{2}\sin x$.

Step2: Apply the product - rule

The product - rule states that if $y = u\cdot v$, where $u = 5x^{2}$ and $v=\sin x$, then $y'=u'v + uv'$. The derivative of $u = 5x^{2}$ is $u'=10x$ (using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$), and the derivative of $v=\sin x$ is $v'=\cos x$.

Step3: Calculate the derivative

$f'(x)=u'v+uv'=(10x)\sin x+5x^{2}\cos x = 10x\sin x + 5x^{2}\cos x$.

Answer:

$10x\sin x+5x^{2}\cos x$