By combining elementary functions and rules of derivation that you already know, create new functions (at least 3) and derive them:
a) $$f(x)=2x^3\tan(x)+\cos(x) \cdot e^x$$
b) $$f(x)=e^x \ln(x)-(5x^2-x^3) \cdot \cos(x)$$
c)$$f(x)=x^3$$ (use the rule of the product)
Development:
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: $$$f(x)=x\cdot x^2 \Rightarrow f'(x)=1\cdot x^2+x\cdot 2x=3x^2$$$
Solution:
a) $$f'(x)=6x^2\tan(x)+2x^3\dfrac{1}{\cos^2(x)}+e^x(\cos(x)-\sin(x))$$
b) $$f'(x)=e^x(\ln(x)+\dfrac{1}{x})-(10x-3x^2)\cos(x)+(5x^2-x^3)\sin(x)$$
c) $$f'(x)=3x^2$$