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Hummingbird 45
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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:
"What is the slope of the line tangent to the graph of the function [itex]f(x)=\ln (\sin^2 x+3)[/itex] at the point where [itex]x=\frac{\pi}{3}[/itex]?"
The correct answer is given as [itex]\frac{2\sqrt{3}}{15}[/itex]
The relevant equation is given above.
No matter how many times I calculate it out, I keep getting [itex]f'(\frac{\pi}{3})=\frac{45\sqrt{3}}{8\pi}[/itex]. 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.)
[tex]f(x)=\ln ((\sin x)^2+3)[/tex]
[tex]f'(x)=\frac{1}{x} ((\sin x)^2+3)\cdot 2\sin x\cdot \cos x[/tex]
[tex]f'(x)=\frac{((\sin x)^2+3)}{x} \cdot 2\sin x\cdot \cos x[/tex]
[tex]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}[/tex]
[tex]f'(\frac{\pi}{3})=\frac{((\frac{\sqrt{3}}{2})^2+3)}{\frac{\pi}{3}} \cdot 2\frac{\sqrt{3}}{2}\cdot \frac{1}{2}[/tex]
[tex]f'(\frac{\pi}{3})=(\frac{3}{4}+3) \cdot \frac{3}{\pi} \cdot \frac{2\sqrt{3}}{4}[/tex]
[tex]f'(\frac{\pi}{3})=(\frac{3}{4}+\frac{12}{4}) \cdot \frac{3}{\pi} \cdot \frac{2\sqrt{3}}{4}[/tex]
[tex]f'(\frac{\pi}{3})=\frac{15}{4} \cdot \frac{3}{\pi} \cdot \frac{2\sqrt{3}}{4}[/tex]
[tex]f'(\frac{\pi}{3})=\frac{15}{4} \cdot \frac{3}{\pi} \cdot \frac{\sqrt{3}}{2}[/tex]
[tex]f'(\frac{\pi}{3})=\frac{45}{4\pi} \cdot \frac{\sqrt{3}}{2}[/tex]
[tex]f'(\frac{\pi}{3})=\frac{45\sqrt{3}}{8\pi}[/tex]
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 [itex]f(x)=\ln (\sin^2 x+3)[/itex] at the point where [itex]x=\frac{\pi}{3}[/itex]?"
The correct answer is given as [itex]\frac{2\sqrt{3}}{15}[/itex]
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 [itex]f'(\frac{\pi}{3})=\frac{45\sqrt{3}}{8\pi}[/itex]. 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.)
[tex]f(x)=\ln ((\sin x)^2+3)[/tex]
[tex]f'(x)=\frac{1}{x} ((\sin x)^2+3)\cdot 2\sin x\cdot \cos x[/tex]
[tex]f'(x)=\frac{((\sin x)^2+3)}{x} \cdot 2\sin x\cdot \cos x[/tex]
[tex]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}[/tex]
[tex]f'(\frac{\pi}{3})=\frac{((\frac{\sqrt{3}}{2})^2+3)}{\frac{\pi}{3}} \cdot 2\frac{\sqrt{3}}{2}\cdot \frac{1}{2}[/tex]
[tex]f'(\frac{\pi}{3})=(\frac{3}{4}+3) \cdot \frac{3}{\pi} \cdot \frac{2\sqrt{3}}{4}[/tex]
[tex]f'(\frac{\pi}{3})=(\frac{3}{4}+\frac{12}{4}) \cdot \frac{3}{\pi} \cdot \frac{2\sqrt{3}}{4}[/tex]
[tex]f'(\frac{\pi}{3})=\frac{15}{4} \cdot \frac{3}{\pi} \cdot \frac{2\sqrt{3}}{4}[/tex]
[tex]f'(\frac{\pi}{3})=\frac{15}{4} \cdot \frac{3}{\pi} \cdot \frac{\sqrt{3}}{2}[/tex]
[tex]f'(\frac{\pi}{3})=\frac{45}{4\pi} \cdot \frac{\sqrt{3}}{2}[/tex]
[tex]f'(\frac{\pi}{3})=\frac{45\sqrt{3}}{8\pi}[/tex]
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