Question

need help on the attachment

Answer 1

\[y=\sqrt{x}e^{x^2-x}(x+1)^{\frac{3}{2}}\]right?

Answer 2

take the log, rewrite using properties of the log, find the derivative, and then multiply back by the original funciton

Answer 3

wrong problem I meant to put up different problem, sorry

Answer 4

\[\ln(\sqrt{x}e^{x^2-x}(x+1)^{\frac{3}{2}}\]
\[\ln(\sqrt{x})+\ln(e^{x^2-x})+\frac{3}{2}\ln(x+1)\]

Answer 5

oops ok

Answer 6

That the actually problem, the one you solve I knew how to do already

Answer 7

ok well this is rather a silly problem but we can do it i think
\[\frac{x}{|x|}=1\text { if } x > 0\] and
\[\frac{x}{|x|}=-1\text { if } x < 0\]
use
\[g(x)=x^2,f(x)=\sqrt{x}\] and the chain rule to find the derivative of
\[f\circ g(x)=\sqrt{x^2}\] you get
\[(f\circ g(x))'=f'(g(x))\times g'(x)=\frac{1}{2\sqrt{x^2}}\times 2x=\frac{x}{\sqrt{x^2}}=\frac{x}{|x|}\]

Answer 8

what there was a problem of finding the derivative of g(x) =sin⁡|x| or f(x) = |sin⁡x|

Answer 9

you mean by the chain rule?

Answer 10

yeah

Answer 11

ok well for
\[f(x)=|\sin(x)|\] then you should really say
\[f(x) = |\sin(x)| = \left\{\begin{array}{rcc}
\sin(x) & \text{if} & \sin(x) \geq 0 \\
-\sin(x)& \text{if} & \sin(x) < 0
\end{array}
\right.
\]

Answer 12

so the derivative will be
\[\cos(x)\] or
\[-\cos(x)\] depending on the interval

Answer 13

but if you want to do it as the chain rule using what you have above, you could say
\[f'(x)=\frac{\sin(x)\cos(x)}{|\sin(x)|}\] which really just says the same thing

Answer 14

since
\[\frac{\sin(x)}{|\sin(x)|}\] is either 1 or -1

Answer 15

why 1 or -1. are you dividing sin by sin

Answer 16

right. it is either 1 or -1 because |sin(x)| is either sin(x) or - sin(x) depending on whether sine is positive or negative

Answer 17

so if sin(x) is positive then sin(x)/|sin(x)|=1
and if sin(x) is negative then sin(x)/|sin(x)|=-1

Answer 18

ok, so what about If g(x) =sin⁡|x|

Answer 19

then by the chain rule you get
\[g'(x)=\cos(|x|)\times \frac{x}{|x|}\] and again the second part is either 1 or -1 depending on whether x is positive or negative

Answer 20

so how would the combine form of that look like?

Answer 21

not sure what you mean

Answer 22

What happen to the cosine on the numerator for last problem you did

Answer 23

you mean
\[g'(x)\]?

Answer 24

f'(X)

Answer 25

it is
\[g'(x)=\frac{x\cos(|x|)}{|x|}\] of

Answer 26

\[f'(x)=\frac{\sin(x)\cos(x)}{|\sin(x)|}\]the cosine is still there. so it is either cosine or - cosine depending on whether sine is positive or negative

Answer 27

oh, ok thanks again satellite

Answer 28

yw