Derivative of exponential, logarithmic and a pow x function

a) I identify two functions that are added: $2x^3\tan(x)$ and $\cos(x)e^x$

Rule of the sum: I must add the derivative of these two functions ('the derivative of the sum is the sum of the derivatives').

Rule of the product: to calculate the derivative of two functions I use the rule of the product.

Let's derive step by step:

$$2x^3\tan(x) \Rightarrow 6x^2\tan(x)+2x^3\dfrac{1}{\cos^2(x)}$$

$$\cos(x)e^x \Rightarrow -\sin(x)e^x+\cos(x)e^x=e^x(\cos(x)-\sin(x))$$

Therefore,

$$f'(x)=6x^2\tan(x)+2x^3\dfrac{1}{\cos^2(x)}+e^x(\cos(x)-\sin(x))$$

b) We must derive the two components of the sum and then add up

$$e^x\ln(x) \Rightarrow e^x\ln(x)+e^x\dfrac{1}{x}$$

$$(5x^2-x^3)\cos(x) \Rightarrow (10x-3x^2)\cos(x)+(5x^2-x^3)(-sin(x))$$

Therefore,

$$f'(x)=e^x(\ln(x)+\dfrac{1}{x})-(10x-3x^2)\cos(x)+(5x^2-x^3)\sin(x)$$

c) We must obtain $x^3$ as a product of two functions: $f(x)=g(x) h(x)$

I identify $g(x)=x$ and $h(x)=x^2$

We use the rule of the product: $$f(x)=x\cdot x^2 \Rightarrow f'(x)=1\cdot x^2+x\cdot 2x=3x^2$$