Problems from Derivative of exponential, logarithmic and a pow x function

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)

See development and solution

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) $$f'(x)=3x^2$$

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