Question

question 1
2 pts
differentiate (f(x)=e^{x}\tan x)
(e^{x}(sec^{2}x + \tan x))(xe^{x - 1}sec^{2}x)(\frac{e^{x}cos x - e^{x}sin x}{cos^{2}x})(e^{x}sec^{2}x)
question 2
2 pts
differentiate (y = xsin x+sec x)(xcos x+sec^{2}x)(sin x+sec x\tan x)(xcos x+sin x+sec x\tan x)(xcos x-sin x+sec x\tan x)

Explanation:

Step1: Apply product - rule for Question 1

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

Step2: Apply sum - rule and product - rule for Question 2

The sum - rule states that if $y = u + w$, then $y^\prime=u^\prime+w^\prime$. For $y=x\sin x+\sec x$, let $u = x\sin x$ and $w=\sec x$.
For $u = x\sin x$, using the product - rule ($u = a\cdot b$ where $a = x$ and $b=\sin x$, $\frac{d}{dx}(a)=1$, $\frac{d}{dx}(b)=\cos x$), we have $u^\prime=\sin x + x\cos x$. And $\frac{d}{dx}(\sec x)=\sec x\tan x$. So $y^\prime=\sin x + x\cos x+\sec x\tan x$.

Answer:

Question 1: A. $e^{x}(\sec^{2}x+\tan x)$
Question 2: C. $x\cos x+\sin x+\sec x\tan x$