Problems from Derivative of trigonometric functions

Development:

a) Using the rule of the sum, we recognize $g(x)=x$ and $h(x)=\sin (x)$, and therefore, $$f^{\prime }(x)=1+\cos (x)$$

b) In this case, $g(x)=5\cos (x)$ and $h(x)=16$. Therefore, $$f^{\prime }(x)=5(-\sin (x))+16(2x)=32x-5\sin (x)$$

c) We recognize now three different functions; we apply the rule of the sum and obtain: $$f^{\prime }(x)={\displaystyle \frac{1}{1+x^2}}-\sin (x)-6x^5$$

d) By applying the rule of the sum: $$f^{\prime }(x)=-{\csc }^2(x)-(-\csc (x)\cot (x))=\csc (x)(\cot (x)-\csc (x))$$

Solution:

a) $f^{\prime }(x)=1+\cos (x)$

b) $f^{\prime }(x)=32x-5\sin (x)$

c) $f^{\prime }(x)={\displaystyle \frac{1}{1+x^2}}-\sin (x)-6x^5$

d) $f^{\prime }(x)=csc(x)(cot(x)-csc(x))$

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