Question

inverse trig, exponentials & logarithms
1 calculate the following derivatives.
a. $\frac{d}{dx}3^{sin 3x}$
b. $\frac{d}{dx}ln(3x^{2}+1)$
c. $\frac{d}{dx}arctan(3x^{2})$
d. $\frac{d}{dx}(x^{pi}cdotpi^{x})$
e. $\frac{d}{dx}sqrt{e^{5x}-7^{x}+ln x}$
f. $\frac{d}{dx}(2^{x}cdotlog_{2}(x))$
g. $\frac{d}{dt}(e^{2t}cdotsin^{-1}(t^{3}))$
h. $\frac{d}{dw}(\frac{ln(w^{2})+e^{5}}{\tan^{-1}(w)})$
i. $\frac{d}{dx}ln(ln(ln(x)))$

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. \(\frac{1}{x\ln x\ln(\ln x)}\)