Finding the tangent line at pi/3 for the following equation

[Hummingbird 45]

Hello All,

I have been trying to brush up on some calculus (differential, I haven't learned integral yet) on my own by finding whatever calculus problems I can find. Recently I found this question listed as a CLEP practice question, and have been having some difficulty with it. Here is the question:

Homework Statement

"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=\frac{\pi}{3}?"

The correct answer is given as \frac{2\sqrt{3}}{15}

Homework Equations

The relevant equation is given above.

The Attempt at a Solution

No matter how many times I calculate it out, I keep getting f'(\frac{\pi}{3})=\frac{45\sqrt{3}}{8\pi}. Here is the work I have done. The error could be anywhere; in the algebra, trig, or calc methodology (math has never been my strongest subject.)

f(x)=\ln ((\sin x)^2+3)
f'(x)=\frac{1}{x} ((\sin x)^2+3)\cdot 2\sin x\cdot \cos x
f'(x)=\frac{((\sin x)^2+3)}{x} \cdot 2\sin x\cdot \cos x
f'(\frac{\pi}{3})=\frac{((\sin \frac{\pi}{3})^2+3)}{\frac{\pi}{3}} \cdot 2\sin \frac{\pi}{3}\cdot \cos \frac{\pi}{3}
f'(\frac{\pi}{3})=\frac{((\frac{\sqrt{3}}{2})^2+3)}{\frac{\pi}{3}} \cdot 2\frac{\sqrt{3}}{2}\cdot \frac{1}{2}
f'(\frac{\pi}{3})=(\frac{3}{4}+3) \cdot \frac{3}{\pi} \cdot \frac{2\sqrt{3}}{4}
f'(\frac{\pi}{3})=(\frac{3}{4}+\frac{12}{4}) \cdot \frac{3}{\pi} \cdot \frac{2\sqrt{3}}{4}
f'(\frac{\pi}{3})=\frac{15}{4} \cdot \frac{3}{\pi} \cdot \frac{2\sqrt{3}}{4}
f'(\frac{\pi}{3})=\frac{15}{4} \cdot \frac{3}{\pi} \cdot \frac{\sqrt{3}}{2}
f'(\frac{\pi}{3})=\frac{45}{4\pi} \cdot \frac{\sqrt{3}}{2}
f'(\frac{\pi}{3})=\frac{45\sqrt{3}}{8\pi}

 

[Student100]

f'(x)=\frac{1}{x} ((\sin x)^2+3)\cdot 2\sin x\cdot \cos x

Check how you're calculating that step; perhaps you would do better to use a u sub here.

u = (sin(x))^2 + 3)

 

[Hummingbird 45]

Check how you're calculating that step; perhaps you would do better to use a u sub here.

u = (sin(x))^2 + 3)

Aha! Thank you. So from there, I get:

f'(x) = \frac {1}{\sin^2x+3} \cdot 2 \sin x \cdot \cos x
f'(\frac{\pi}{3}) = \frac {2 \cdot \frac {\sqrt3} {2} \cdot \frac {1}{2}}{\frac{3}{4}+\frac{3}{1}}
f'(\frac{\pi}{3}) = \frac {\frac {\sqrt3} {2}}{\frac{3}{4}+\frac{12}{4}}
f'(\frac{\pi}{3}) = \frac {\frac {\sqrt3} {2}}{\frac{15}{4}}
f'(\frac{\pi}{3}) = \frac {\sqrt3} {2} \cdot \frac{4}{15}
f'(\frac{\pi}{3}) = \frac {2\sqrt3}{15}

 

[Mark44]

Aha! Thank you. So from there, I get:

f'(x) = \frac {1}{\sin^2x+3} \cdot 2 \sin x \cdot \cos x
f'(\frac{\pi}{3}) = \frac {2 \cdot \frac {\sqrt3} {2} \cdot \frac {1}{2}}{\frac{3}{4}+\frac{3}{1}}
f'(\frac{\pi}{3}) = \frac {\frac {\sqrt3} {2}}{\frac{3}{4}+\frac{12}{4}}
f'(\frac{\pi}{3}) = \frac {\frac {\sqrt3} {2}}{\frac{15}{4}}
f'(\frac{\pi}{3}) = \frac {\sqrt3} {2} \cdot \frac{4}{15}
f'(\frac{\pi}{3}) = \frac {2\sqrt3}{15}

You can save yourself some writing by not starting each equation with f'(∏/3). As you simplify each expression, just connect the new expression with =.
$$f'(\frac{\pi}{3}) = \frac {2 \cdot \frac {\sqrt3} {2} \cdot \frac {1}{2}}{\frac{3}{4}+\frac{3}{1}} = \frac {\frac {\sqrt3} {2}}{\frac{3}{4}+\frac{12}{4}}$$
$$ = \frac {\frac {\sqrt3} {2}}{\frac{15}{4}}$$
and so on.

 

[Hummingbird 45]

You can save yourself some writing by not starting each equation with f'(∏/3). As you simplify each expression, just connect the new expression with =.
$$f'(\frac{\pi}{3}) = \frac {2 \cdot \frac {\sqrt3} {2} \cdot \frac {1}{2}}{\frac{3}{4}+\frac{3}{1}} = \frac {\frac {\sqrt3} {2}}{\frac{3}{4}+\frac{12}{4}}$$
$$ = \frac {\frac {\sqrt3} {2}}{\frac{15}{4}}$$
and so on.

Ok, I'll remember that.