Question

Find each of the following derivatives. Simplify all answers completely. Show all work toward your answer or provide an explanation for how you arrived at your answer.

1) f(x)=sqroot e^2x + 8x^2 e^x

2) g(x)= xsin(x)/1+cos(x)

3) h(x)=e^-x sin(x)

4) y=sec(x)tan(x)

5) f(t)= ((t-1)(2t^2 -1))/(t^3 -1)

Answer 1

1)

$$f'(x)=(\sqrt{e^{2x} + 8x^2 e^x})'=\dfrac{(e^{2x} + 8x^2 e^x)'}{2\sqrt{e^{2x} + 8x^2 e^x}}$$

$$=\dfrac{2e^{2x}+16xe^x+8x^2e^x}{2\sqrt{e^{2x} + 8x^2 e^x}}$$

$$=\dfrac{e^{2x}+8xe^x+4x^2e^x}{\sqrt{e^{2x} + 8x^2 e^x}}$$

$$=\dfrac{e^x(e^{x}+8x+4x^2)}{\sqrt{e^{2x} + 8x^2 e^x}}$$

2)

$$g'(x)=(\dfrac{x\sin(x)}{1+\cos(x)})'=(\dfrac{x(2\sin(x/2)\cos(x/2))}{2\cos^2(x/2)})'$$

$$=(x\tan(x/2))'=\tan(x/2)+\dfrac{x}{2\cos^2(x/2)}$$

$$=\dfrac{2\sin(x/2)\cos(x/2)+x}{2\cos^2(x/2)}=\dfrac{\sin(x)+x}{1+\cos(x)}$$

3)

$$h'(x)=(e^{-x}\sin(x))'=-e^{-x}\sin(x)+e^{-x}\cos(x)$$

4)

$$y'=(\sec(x)\tan(x))'=(\dfrac{\sin(x)}{\cos^2(x)})'$$

$$=\dfrac{\cos^3(x)-2\cos(x)(-\sin(x))\sin(x)}{\cos^4(x)}$$

$$=\dfrac{\cos^2(x)+2\sin^2(x)}{\cos^3(x)}=\dfrac{1+\sin^2(x)}{\cos^3(x)}$$

$$=\sec^3(x)+\sec(x)\tan^2(x)$$

5)

$$f'(t)=(\dfrac{(t-1)(2t^2-1)}{t^3-1})'=(\dfrac{(t-1)(2t^2-1)}{(t-1)(t^2+t+1)})'$$

$$=(\dfrac{2t^2-1}{t^2+t+1})'=\dfrac{4t(t^2+t+1)-(2t+1)(2t^2-1)}{(t^2+t+1)^2}$$

$$=\dfrac{4t^3+4t^2+4t-4t^3+2t-2t^2+1}{(t^2+t+1)^2}$$

$$=\dfrac{2t^2+6t+1}{(t^2+t+1)^2}$$