inverse trig, exponentials & logarithms\n1 calculate the following derivatives.\na. $\frac{d}{dx}3^{sin…

Explanation:

Step1: Recall chain - rule and derivative formulas

The chain - rule states that if (y = f(g(x))), then (y^\prime=f^\prime(g(x))\cdot g^\prime(x)). Also, (\frac{d}{dx}a^x=a^x\ln a), (\frac{d}{dx}\ln x=\frac{1}{x}), (\frac{d}{dx}\arctan x=\frac{1}{1 + x^2}), (\frac{d}{dx}\sin^{-1}x=\frac{1}{\sqrt{1 - x^2}}).

Step2: Solve part A

Let (y = 3^{\sin3x}). Let (u=\sin3x), then (y = 3^{u}). (\frac{dy}{du}=3^{u}\ln3) and (\frac{du}{dx}=3\cos3x). By the chain - rule, (\frac{d}{dx}3^{\sin3x}=3^{\sin3x}\ln3\cdot3\cos3x = 3\ln3\cdot3^{\sin3x}\cos3x).

Step3: Solve part B

Let (y=\ln(3x^{2}+1)). Let (u = 3x^{2}+1), then (y=\ln u). (\frac{dy}{du}=\frac{1}{u}) and (\frac{du}{dx}=6x). By the chain - rule, (\frac{d}{dx}\ln(3x^{2}+1)=\frac{6x}{3x^{2}+1}).

Step4: Solve part C

Let (y = \arctan(3x^{2})). Let (u = 3x^{2}), then (y=\arctan u). (\frac{dy}{du}=\frac{1}{1 + u^{2}}) and (\frac{du}{dx}=6x). By the chain - rule, (\frac{d}{dx}\arctan(3x^{2})=\frac{6x}{1 + 9x^{4}}).

Step5: Solve part D

Using the product - rule ((uv)^\prime = u^\prime v+uv^\prime), where (u = x^{\pi}) and (v=\pi^{x}). (\frac{d}{dx}x^{\pi}=\pi x^{\pi - 1}) and (\frac{d}{dx}\pi^{x}=\pi^{x}\ln\pi). So (\frac{d}{dx}(x^{\pi}\cdot\pi^{x})=\pi x^{\pi - 1}\cdot\pi^{x}+x^{\pi}\cdot\pi^{x}\ln\pi=\pi^{x + 1}x^{\pi - 1}+x^{\pi}\pi^{x}\ln\pi).

Step6: Solve part E

Let (y=\sqrt{e^{5x}-7x+\ln x}=(e^{5x}-7x+\ln x)^{\frac{1}{2}}). Let (u = e^{5x}-7x+\ln x), then (y = u^{\frac{1}{2}}). (\frac{dy}{du}=\frac{1}{2\sqrt{u}}) and (\frac{du}{dx}=5e^{5x}-7+\frac{1}{x}). By the chain - rule, (\frac{d}{dx}\sqrt{e^{5x}-7x+\ln x}=\frac{5e^{5x}-7+\frac{1}{x}}{2\sqrt{e^{5x}-7x+\ln x}}).

Step7: Solve part F

Using the product - rule ((uv)^\prime = u^\prime v+uv^\prime), where (u = 2^{x}) and (v=\log_{2}x). (\frac{d}{dx}2^{x}=2^{x}\ln2) and (\frac{d}{dx}\log_{2}x=\frac{1}{x\ln2}). So (\frac{d}{dx}(2^{x}\cdot\log_{2}x)=2^{x}\ln2\cdot\log_{2}x+\frac{2^{x}}{x\ln2}).

Step8: Solve part G

Using the product - rule ((uv)^\prime = u^\prime v+uv^\prime), where (u = e^{2t}) and (v=\sin^{-1}(t^{3})). (\frac{d}{dt}e^{2t}=2e^{2t}) and (\frac{d}{dt}\sin^{-1}(t^{3})=\frac{3t^{2}}{\sqrt{1 - t^{6}}}). So (\frac{d}{dt}(e^{2t}\cdot\sin^{-1}(t^{3}))=2e^{2t}\sin^{-1}(t^{3})+\frac{3t^{2}e^{2t}}{\sqrt{1 - t^{6}}}).

Step9: Solve part H

Using the quotient - rule ((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}), where (u=\ln(w^{2})+e^{5}) and (v = \tan^{-1}(w)). (u^\prime=\frac{2}{w}) and (v^\prime=\frac{1}{1 + w^{2}}). So (\frac{d}{dw}(\frac{\ln(w^{2})+e^{5}}{\tan^{-1}(w)})=\frac{\frac{2}{w}\tan^{-1}(w)-\frac{\ln(w^{2}) + e^{5}}{1 + w^{2}}}{(\tan^{-1}(w))^{2}}).

Step10: Solve part I

Let (y=\ln(\ln(\ln x))). Let (u=\ln(\ln x)), then (y = \ln u). (\frac{dy}{du}=\frac{1}{u}) and (\frac{du}{dx}=\frac{1}{x\ln x}). By the chain - rule, (\frac{d}{dx}\ln(\ln(\ln x))=\frac{1}{x\ln x\ln(\ln x)}).

Answer:

A. (3\ln3\cdot3^{\sin3x}\cos3x) B. (\frac{6x}{3x^{2}+1}) C. (\frac{6x}{1 + 9x^{4}}) D. (\pi^{x + 1}x^{\pi - 1}+x^{\pi}\pi^{x}\ln\pi) E. (\frac{5e^{5x}-7+\frac{1}{x}}{2\sqrt{e^{5x}-7x+\ln x}}) F. (2^{x}\ln2\cdot\log_{2}x+\frac{2^{x}}{x\ln2}) G. (2e^{2t}\sin^{-1}(t^{3})+\frac{3t^{2}e^{2t}}{\sqrt{1 - t^{6}}}) H. (\frac{\frac{2}{w}\tan^{-1}(w)-\frac{\ln(w^{2}) + e^{5}}{1 + w^{2}}}{(\tan^{-1}(w))^{2}}) I. (\frac{1}{x\ln x\ln(\ln x)})