Answer to Question #198361 in Calculus for Zinzi Nkonco

Question

Determine the derivatives of the following functions:

a.f(x)=In(x+Inx)

b.g(x)=In√x-1÷x^4+1

C.h(x)=√xe^x^2-x(x+1)^2/3

d.y(x)=(sinx)Inx

Determine the derivatives of the following inverse trigonometric functions:

a.f(x)=tan^-1√x

b.y(x)=In(x^2cot^-1x/√x-1)

c.g(x)=sin^-1(3x)+cos^-1(x/2)

d.h(x)=tan-1(x-√x^2+1)

Answer 1

$I.\newline a.\newline f(x)=In(x+Inx)\newline f'=\frac{1}{x+lnx}(1+\frac{1}{x})\newline b.\newline g=\frac{ln\sqrt{x-1}}{x^4+1}\newline g'=\frac{(x^2+1)(\frac{1}{\sqrt{x-1}})(\frac{1}{2\sqrt{x-1}})-(ln\sqrt{x-1})(2x)}{(x^2+1)^2}\newline =\frac{x^2+1-4x(x-1)ln\sqrt{x-1}}{2(x-1)(x^2+1)^2} \newline c.\newline h(x)=\sqrt{xe^{x^2}-x(x+1)^{\frac{2}{3}}}\newline h'=\frac{1}{\sqrt{xe^{x^2}-x(x+1)^{\frac{2}{3}}}}(xe^{x^2}(2x)+e^{x^2}-x(\frac{2}{3}(x+1)^{\frac{-1}{3}})-(x+1)^{\frac{2}{3}})\newline =\frac{1}{\sqrt{xe^{x^2}-x(x+1)^{\frac{2}{3}}}}(e^{x^2}(2x^2+1)-\frac{2}{3}x(x+1)^{\frac{-1}{3}})-(x+1)^{\frac{2}{3}})\newline d.\newline y(x)=(sinx)lnx\newline y'=sinx(\frac{1}{x})+lnx(cosx)\newline II.\newline a.\newline f=tan^{-1}\sqrt{x}\newline f'=\frac{1}{1+(\sqrt{x})^2}\frac{1}{2\sqrt{x}}\newline =\frac{1}{2(1+x)\sqrt{x}}\newline b.\newline y(x)=In(x^2cot^{-1}(\frac{x}{\sqrt{x-1}}))\newline y'=\frac{1}{x^2cot^{-1}(\frac{x}{\sqrt{x-1}})}.(x^2(\frac{-1}{1+(\frac{x}{\sqrt{x-1}})^2})+cot^{-1}(\frac{x}{\sqrt{x-1}})(2x))\newline =\frac{1}{x^2cot^{-1}(\frac{x}{\sqrt{x-1}})}.(\frac{-x^2}{x^2+x+1}+2xcot^{-1}(\frac{x}{\sqrt{x-1}})) \newline c.\newline g(x)=sin^{-1}(3x)+cos^{-1}(\frac{x}{2})\newline g'=\frac{1}{\sqrt{1-(3x)^2}}(3)+\frac{-1}{\sqrt{1-(\frac{x}{2})}}\frac{1}{2}\newline =\frac{3}{\sqrt{1-9x^2}}-\frac{1}{\sqrt{4-2x}}\newline d.\newline h(x)=tan^{-1}(x-\sqrt{x^2+1})\newline h'=\frac{1}{1+(x-\sqrt{x^2+1})^2}.(1-\frac{1}{2\sqrt{x^2+1}}(2x)\newline =\frac{1}{1+(x-\sqrt{x^2+1})^2}.(\frac{\sqrt{x^2+1}-x}{\sqrt{x^2+1}})$