Question

Find dy/dx of y=x(lnx)^(sin^2x)

using log differentiation.

lny = ln x + (sin^2 x) ln (lnx)

y`/y = 1/x + sin2x (ln(lnx)) + sin^2x/(x lnx)

Can someone explain to be thoroughly the

lny = ln x + (sin^2 x) ln (lnx)

step please!!!!! Test tomorrow

Answer 1

y=x(lnx)^(sin^2x)

take ln both sides, what do you get ?

Answer 2

lny = .... ln(x(lnx)^(sin^2x))

Answer 3

But wouldn't that =

Answer 4

lny = (sin^2x)ln(x)(lnx) ? I'm messing up badly on this step lol

Answer 5

$$
\Large y=x~ln(x)^{sin^2x} \\
\text{take ln of both sides} \\
\Large ln(y) = ln ( x~ln(x)^{sin^2x})

$$

Answer 6

My eyes tell me that this is the log(a*b) rule but I don't see that in the answer either

Answer 7

$$
\Large y=x\cdot ln(x)^{sin^2x} \\
\text{take ln of both sides} \\
\Large ln(y) = ln ( x\cdot ln(x)^{sin^2x}) = ln ( x) + ln( ln(x)^{sin^2x})

$$

Answer 8

Oh okay I see the mult rule now, cool beans

Answer 9

And then the power is pulled down afterwards

Answer 10

$$
\Large \color{blue} {\text{Log rules:}}\\
\Large \log(A \cdot B) = \log(A)+ ln(B)\\
\Large \log(\frac{A}{B}) = \log(A)- ln(B)\\
\Large \log(A^n) = n\cdot \log(A)

$$

Answer 11

Thank ya this helps me. I have a couple more questions to clear up for my test if you want to have some more calc fun look for my questions! :D

Answer 12

sure :)