Answer to Question #223106 in Calculus for Haddensen

Question

Find f'(x) iff(x) = e^x(1+Inx)

f(x) = x³e^1/x

f(x) = √[x²-1](x+1) / (x²+1)^3/2

Answer 1

• Solution: $f(x)=e^x (1+ln(x))$

By the Product Rule, we have

$$f'(x)=\frac{d}{dx}(e^x(1+ln(x)))= (1+ln(x))\frac{d}{dx}e^x$$ $$+e^x \frac{d}{dx}(1+ln(x))=\{the\space Sum \space Rule\}$$ $$=(1+ln(x))\frac{d}{dx}e^x+e^x (\frac{d}{dx}1+\frac{d}{dx}ln(x))$$ $$=(1+ln(x))e^x+e^x (0+\frac{1}{x})=e^x(1+ln(x)+\frac{1}{x})$$

Answer: $f'(x)=e^x (1+ln(x)+\frac{1}{x})$

• Solution: $f(x)=x^3e^{1/x}$

By the Product Rule, we have

$$f'(x)=\frac{d}{dx}(x^3e^{1/x} )=e^{1/x}\frac{d}{dx}x^3+x^3 \frac{d}{dx}e^{1/x}$$

By the Chain Rule: let u=1/x

$$f'(x)=e^{1/x}\frac{d}{dx}x^3+x^3 \frac{d}{du}e^u\frac{d}{dx}(1/x)$$ $$=e^{1/x}3x^2+x^3 e^{1/x}(-1/x^2)=xe^{1/x}(3x-1)$$

Answer: $f'(x)=xe^{1/x}(3x-1)$

• Solution: $f(x)=\frac{(x+1)\sqrt{x^2 -1}}{(x^2+1)^{3/2}}$

By the Quotient Rule, we have

$$f'(x)=\frac{d}{dx}\Big(\frac{(x+1)\sqrt{x^2 -1}}{(x^2+1)^{3/2}} \Big)$$ $$=\frac{(x^2+1)^{3/2}\frac{d}{dx}((x+1)\sqrt{x^2 -1})-(x+1)\sqrt{x^2 -1}\frac{d}{dx}(x^2+1)^{3/2}}{((x^2+1)^{3/2})^2} \space (1)$$

By the Product Rule

$$\frac{d}{dx}((x+1)\sqrt{x^2 -1})=\sqrt{x^2 -1}\frac{d}{dx}(x+1)$$ $$+(x+1)\frac{d}{dx}\sqrt{x^2 -1} \qquad (2)$$

By the Sum Rule

$$\frac{d}{dx}(x+1)=\frac{d}{dx}x+\frac{d}{dx}1=1+0=1 \qquad (3)$$

By the Chain Rule: let $u=x^2-1$

$$\frac{d}{dx}\sqrt{x^2 -1}=\frac{d}{du}\sqrt{u}\frac{d}{dx}(x^2 -1)=\{the\space Difference \space Rule\}$$ $$=\frac{d}{du}(\sqrt{u})\Big(\frac{d}{dx}x^2 -\frac{d}{dx}1\Big)=\frac{1}{2\sqrt{x^2 -1}}(2x-0)$$ $$=\frac{x}{\sqrt{x^2 -1}}\qquad(4)$$

substitute (3), (4) into (2)

$$\frac{d}{dx}((x+1)\sqrt{x^2 -1})=\sqrt{x^2 -1}+\frac{x(x+1)}{\sqrt{x^2 -1}}\qquad (5)$$

By the Chain Rule: let $u=x^2+1$

$$\frac{d}{dx}(x^2+1)^{3/2}=\frac{du^{3/2}}{du}\frac{d}{dx}(x^2+1)=\frac{du^{3/2}}{du}(\frac{d}{dx}x^2+\frac{d}{dx}1)$$ $$=\frac 3 2 \sqrt{x^2 +1}(2x+0)=3x\sqrt{x^2 +1}\qquad (6)$$

substitute (5), (6) into (1)

$$f'(x)=\frac{(x^2+1)^{3/2}(\sqrt{x^2 -1}+\frac{x(x+1)}{\sqrt{x^2 -1}})-3x(x+1)\sqrt{x^2 -1}\sqrt{x^2 +1}}{(x^2+1)^3}$$ $$=\frac{\sqrt{x^2 -1}+\frac{x(x+1)}{\sqrt{x^2 -1}}}{(x^2+1)^{3/2}}-\frac{3x(x+1)\sqrt{x^2 -1}}{(x^2+1)^{5/2}}$$ $$=\frac{\sqrt{x^2 -1}}{(x^2+1)^{3/2}}+\frac{x(x+1)}{\sqrt{x^2 -1}(x^2+1)^{3/2}}-\frac{3x(x+1)\sqrt{x^2 -1}}{(x^2+1)^{5/2}}$$

Answer: $f'(x)=\frac{\sqrt{x^2 -1}}{(x^2+1)^{3/2}}+\frac{x(x+1)}{\sqrt{x^2 -1}(x^2+1)^{3/2}}-\frac{3x(x+1)\sqrt{x^2 -1}}{(x^2+1)^{5/2}}$