Question

What is the slope of the line tangent to the graph of the function f(x) = ln(sin^2(x)+3) at the point where x=(pi/3)?

Answer 1

I know that this is searching for the derivative and I need to use the chain rule but I get hung up about halfway through.
dy/dx of ln(x) = 1/x
dy/dx of sin^2(x)+3 = 2sin(x)

So... f o g is:
\[\frac{ 1 }{ \sin ^{2}(x)+3 }\times2\sin(x)\]

However, I'm not sure where to go after that.

Answer 2

take derivative using chain rule
\[f'(x) = \frac{2\sin x \cos x}{\sin^{2} x +3}\]

Answer 3

pretty close...but you have to apply chain rule to sin^2 as well ... derivative of sin is cos

Answer 4

after that plug in x = pi/3 ... that number is slope of tangent line

Answer 5

Can you explain why chain rule needs to be applied to sin^2? Isn't f o g just:
\[f \prime (g(x)) \times g \prime (x)?\]

Answer 6

you are correct...but g(x) = sin x

Answer 7

Lol, thank you for the picture. I didn't realize the derivative of sin(x) also needs to be included. Thanks dumbcow as well, you helped me out big time as well.

Answer 8

One more question on this:

do I plug in the values of pi/3 using the unit circle?

In that case cos(pi/3) = \[\frac{ 1 }{ 2 }\]
and sin(pi/3) = \[\frac{ \sqrt{3} }{ 2 }\]

Once those are plugged into the equation i'm a bit confused on how to simplify.. my algebra is rusty (. _ . )

The solution should be \[\frac{ 2\sqrt{3} }{ 15 }\] but I'm not sure how to get there.

Answer 9

\[f'(\frac{\pi}{3}) = \frac{2\frac{\sqrt{3}}{2} \frac{1}{2}}{(\frac{\sqrt{3}}{2})^{2} +3} = \frac{\frac{\sqrt{3}}{2}}{\frac{3}{4}+\frac{12}{4}}=\frac{\sqrt{3}}{2}*\frac{4}{15} =\frac{2\sqrt{3}}{15} \]