Question

Differentiate the function.
f(x) = ln(25 sin^2(x))

I got f'(x)= ((25/2)(x-sin(x)cos(x))/(25 sin^2 (x))

Is that right? If it is, how do I simplify it?

Answer 1

no it is not

Answer 2

you got denominator correct, but not numerator

you have \(25\sin^2x = 25(\sin x)^2\), you can use chain rule here

Answer 3

I would re-write 25sin²x as (5sinx)²

and sub u² instead of 5sin²

Answer 4

So would it be
\[\frac{ 2u }{ u^2 } = \frac{ 2 (5 \sin x) }{ 25 \sin^2 x } = \frac{ 10 \sin x }{ 25 \sin^2 x }\] ?

Answer 5

\(\Large (u^2)' = 2u\cdot u'\)

Answer 6

you have \(\ln (u^2) = \dfrac{(u^2)'}{u^2}=\dfrac{2u\cdot u'}{u^2}\)

Answer 7

given that \(u = 5\sin x\)

Answer 8

Is this right?
\[\ln(u^2) = \frac{ (u^2)' }{ u^2 } = \frac{ 2u \times u' }{ u^2 } = \frac{ 2(5 \sin x)(5 \cos x) }{ 25 \sin^2 x } = \frac{ 50 \sin x \cos x }{ 25 \sin^2 x }\]

Answer 9

Ugh the u's make it so confusing in my opinion.
But yes, very good :) Looks correct.

Answer 10

\(\Large f(x) = \ln(25 \sin^2(x)) = \ln (5\sin(x))^2 = \\ \Large
2*\ln(5\sin(x)) = 2\ln(5) + 2\ln(\sin(x)) \)

\(\Large f'(x) = 2\frac{1}{\sin(x)}*(\cos(x)) = 2\cot(x)\)

Answer 11

@aum

Answer 12

lol

Answer 13

Haha xD
Thanks so much, you guys! :)
There are so many notes in my binder now that it's hard to find what you're looking for.

Answer 14

You are welcome.