Answer to Question #242413 in Calculus for lalaland

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)

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

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

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

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