Finding derivatives using Chain Rule

[lamaclass]

Hi, I am having difficulty with solving these problems using the Chain Rule. I was able to solve them partially but then got stuck. Any help on how to continue or finish them would be great!

Find the derivative of the function:

1. f(x)= ln (x/x[sup:36e5yc0i]2[/sup:36e5yc0i]+1)

f'(x)= ln [x(x[sup:36e5yc0i]2[/sup:36e5yc0i]+1)[sup:36e5yc0i]-1[/sup:36e5yc0i])]

f'(x)= ln[(-x(x[sup:36e5yc0i]2[/sup:36e5yc0i]+1)[sup:36e5yc0i]-2[/sup:36e5yc0i]]*2x

?f'(x)= ln[-2x[sup:36e5yc0i]2[/sup:36e5yc0i](x[sup:36e5yc0i]2[/sup:36e5yc0i]+1)[sup:36e5yc0i]-2[/sup:36e5yc0i]]

Find the second derivative of the function:

2. f(x)= sin x[sup:36e5yc0i]2[/sup:36e5yc0i]

f'(x)= cos x[sup:36e5yc0i]2[/sup:36e5yc0i]*2x
=2x cos x[sup:36e5yc0i]2[/sup:36e5yc0i]

f''(x)= 2x[(2x (-sin x[sup:36e5yc0i]2[/sup:36e5yc0i])+cos x[sup:36e5yc0i]2[/sup:36e5yc0i]*x)]

?f''(x)= 2x[2 cos x[sup:36e5yc0i]2[/sup:36e5yc0i]-2x sin x[sup:36e5yc0i]2[/sup:36e5yc0i]]

Find the derivative of the function:

3. g(t)= t[sup:36e5yc0i]2[/sup:36e5yc0i]2[sup:36e5yc0i]t[/sup:36e5yc0i]

g'(t)= t[sup:36e5yc0i]2[/sup:36e5yc0i](ln 2)2t+2[sup:36e5yc0i]t[/sup:36e5yc0i]*2t

Wasn't sure how to continue on from here.

Find the derivative of the function:

Did not know if I did this one correctly or not.

4. y=log [sub:36e5yc0i]5[/sub:36e5yc0i](x[sup:36e5yc0i]2[/sup:36e5yc0i]+1)[sup:36e5yc0i]1/2[/sup:36e5yc0i]

y'=log [sub:36e5yc0i]5[/sub:36e5yc0i][(1/2(x[sup:36e5yc0i]2[/sup:36e5yc0i]+1)[sup:36e5yc0i]-1/2[/sup:36e5yc0i]]*2x

y'=log [sub:36e5yc0i]5[/sub:36e5yc0i] [x(x[sup:36e5yc0i]2[/sup:36e5yc0i]+1)[sup:36e5yc0i]-1/2[/sup:36e5yc0i])

y'=(x[sup:36e5yc0i]2[/sup:36e5yc0i]+1)[sup:36e5yc0i]1/2[/sup:36e5yc0i]/(ln 5) x

 

[soroban]

Hello, lamaclass!

\(\displaystyle [1]\;\;f(x)\:=\: \ln\left(\frac{x}{x^2+1}\right)\)

We can do this head on . . .

\(\displaystyle f'(x) \;=\;\frac{1}{\frac{x}{x^2+1}} \cdot\left[\frac{(x^2+1)\cdot1 - x\cdot2x}{(x^2+1)^2}\right] \;=\;\frac{x^2+1}{x}\cdot\frac{1-x^2}{(x^2+1)^2} \;=\;\frac{1-x^2}{x(x^2+1)}\)


\(\displaystyle \text{Or we can simplify first: }\;f(x) \;=\;\ln\left(\frac{x}{x^2+1}\right) \;=\;\ln(x) - \ln(x^2+1)\)
. . \(\displaystyle \text{Then: }\;f'(x) \;=\;\frac{1}{x} - \frac{2x}{x^2+1}\quad\hdots\quad\text{same answer}\)

\(\displaystyle \text{[2] Find the second derivative of: }\; f(x)\;=\; \sin(x^2)\)

\(\displaystyle f'(x) \;=\;\cos(x^2)\cdot 2x \;=\;2x\cos(x^2)\)

\(\displaystyle f''(x) \;=\;2x\cdot[-\sin(x^2)]\!\cdot\!2x + 2\cdot\cos(x^2) \;=\;-4x^2\sin(x^2) + 2\cos(x^2)\)

\(\displaystyle \text{[3] Differentiate: }\;g(t)\:= \;t^2\,2^t\)

\(\displaystyle g'(t)\;=\; t^2 \cdot(\ln 2)2^t + 2^t\cdot2t\)

Wasn't sure how to continue on from here.

Note much we can do . . . maybe factor.

\(\displaystyle f'(x) \;=\;t\!\cdot\!2^t\bigg[t\!\cdot\!\ln 2 + 2\bigg]\)

\(\displaystyle \text{[4] Differentiate: }\:y \;=\;\log_5\left(x^2+1\right)^{\frac{1}{2}\)

\(\displaystyle \text{Head-on: }\;y' \;=\;\frac{1}{(x^2+1)^{\frac{1}{2}}}\cdot\frac{1}{2}(x^2+1)^{-\frac{1}{2}}\cdot2x\cdot\frac{1}{\ln 5}\)

. . \(\displaystyle \text{which simplifies to: }\;y' \;=\;\frac{x}{(x^2+1)\cdot\ln 5}\)


\(\displaystyle \text{Simplify first: }\; y \;=\;\tfrac{1}{2}\log_5(x^2+1)\)

\(\displaystyle \text{Then: }\;y' \;=\;\tfrac{1}{2}\cdot\frac{2x}{x+1}\cdot\frac{1}{\ln5} \;=\;\frac{x}{(x^2+1)\xsor\ln5}\)

 

[lamaclass]

Thanks Soroban! I appreciate your help.