Differentiating two trig functions. Check.

Question

I have two problems that I'd like a quick checkup on. I am trying to differentiate these two functions:

1. $f(x) = x \cdot \cos x + 2\cdot \tan x$

and

2. $f(x) = e^x \cdot \cos x$

Here are my attempts:

1. Going to use a combo of sum and product rules.

$$f'(x) = \frac{d}{dx} (x\cdot cosx) + \frac{d}{dx} (2\cdot tanx)$$

$$= x\cdot(-sinx) + cosx + 2sec^2x$$

Is there anyway to simplify from here? It looks like I'm missing something...

2. $$f'(x) = e^x\cdot sinx + cosx \cdot e^x$$

$$e^x(\sin x + \cos x)$$

Here's another check:

3. Differentiate $y = sec\theta \cdot tan\theta$

$$= y' = sec\theta \cdot sec^2\theta + tan\theta \cdot sec\theta \cdot tan\theta$$

$$= sec^3\theta + tan^2\theta \cdot sec\theta$$

$$= sec\theta \cdot (sec^2\theta + tan^2\theta)$$

Answer 1

Perhaps extending the steps will help. We have $$f(x) = x\cos(x) + 2\tan(x)$$

Then

\begin{align*} f'(x) &= \frac{d}{dx}\left(x\cos(x) + 2\tan(x)\right)\\ &= \frac{d}{dx}\bigg( x \bigg) \cos(x) + x\frac{d}{dx}\bigg( \cos(x) \bigg) + \frac{d}{dx}\bigg(2\tan(x) \bigg )\\ &= \cos(x) - x\sin(x) + 2\sec^2(x) \end{align*} For the second we have

$$f(x) = e^{x}\cos(x)$$

Then

\begin{align*} f'(x) &= \frac{d}{dx}\bigg(e^{x}\cos(x)\bigg)\\ &= \frac{d}{dx}\bigg(e^{x}\bigg)\cos(x) + e^{x}\frac{d}{dx}\bigg(\cos(x)\bigg)\\ &= e^{x}\cos(x) - e^{x}\sin(x)\\ &= e^{x}\left(\cos(x) - \sin(x)\right) \end{align*}

Answer 2

First one looks good to me

Third differentiation is correct too.

Correct the second expression

$$(e^x\cos(x))'=e^x(\cos(x)-sin(x))$$