Question

Find the derivative:
h(y)=ln(y^2cosy)
Please show steps, I'm completely lost.

Answer 1

What's your progress?

Answer 2

I have ln(y^2)+ln(cosy)

Answer 3

We can do that by using product rule and some others

Answer 4

OKay

Answer 5

alright after the derivative do you get
\[\large h'(y)=\frac{1}{x^2\cos(x)}(x^2(-\sin(x)+\cos(x)(2x))\]

Answer 6

\[\large h'(y)=\frac{1}{x^2\cos(x)}(x^2(-\sin(x))+\cos(x)(2x))\]

Answer 7

How?

Answer 8

\[y=\ln(u), \ where \ u \ is \ a \ function \ of \ x \ ,then\ y'=\frac{ 1 }{ u }\frac{ du}{ dx }\]

Answer 9

You differentiate ln(x) you will get \(\huge \frac{1}{x}\), same for that, and change those "x" into y's I made a mistake instead of putting it "y"

Answer 10

\[\large h'(y)=\frac{1}{y^2\cos(y)}(y^2(-\sin(y))+\cos(y)(2y))\]

Answer 11

So the way I have it started is incorrect?

Answer 12

Then after differentiating ln(y), you need to differentiate inside of it according to chain rule, you can do that by product rule

Answer 13

You can do that, it doesn't matter, do you get
\[h(y)=\ln(y^2)+\ln(\cos(y)) \\ \\ h'(y)=\frac{1}{y^2}(2y)+\frac{1}{\cos(y)}(-\sin(y))\]

Answer 14

if f(x) =ln(x)
then y' of f(x)=\[\frac{ 1 }{ x } \]

Answer 15

Which way is the correct way?

Answer 16

\[h'(y)=\frac{ 1 }{ y^2\cos(y)}\frac{ d }{ dy }y^2\cos(y)=\frac{ 1 }{ y^2 \cos(y)}2y \cos(y)-y^2\sin(y)=\frac{ 2\cos(y)-y \sin(y) }{ y \cos(y) }\]\[=\frac{ 2 }{ y }-\tan(y)\]
@kaldrid3

Answer 17

Do you see how that works?

Answer 18

sam was right.

Answer 19

I feel like h′(y)=1x2cos(x)(x2(−sin(x))+cos(x)(2x) was the correct answer but now I have three of them, what one is correct?

Answer 20

But why didn't you find the derivative of y?

Answer 21

No that was a mistake you should replace the "x" with "y"

Answer 22

Even so, there is no 2y for the derivative, just y^2 meaning it wasn't derived.

Answer 23

Anyone want to point out which one is correct? I'm seeing the 2/x-tanx as an answer elsewhere too