Question

How can I solve for \(f'(x)\) if \(f(x) = tan(\frac{\pi x}{2})\) and \(0 13 years ago

Answer 1

chain rule

Answer 2

I got \(f'(x) = \frac{cos^2(\frac{\pi x}{2})\frac{\pi}{2}+sin^2(\frac{\pi x}{2})}{cos^2(\frac{\pi x}{2})}\)

Answer 3

how can i simplify that?

Answer 4

well...how u get that? \[(\tan u)'=u' (1+\tan^2 u)\]

Answer 5

tanx = sinx/cosx

Answer 6

oh yeah ... there is a little typo there\[f'(x) =\frac{\pi}{2} \frac{\cos^2(\frac{\pi x}{2})+\sin^2(\frac{\pi x}{2})}{\cos^2(\frac{\pi x}{2})}\]

Answer 7

and for simplifying\[\sin^2 \text{everything}+\cos^2 \text{everything}=1\]:)

Answer 8

Looks like i made a mistake there... the next part of the question is to find the equation of the tangent line to this graph at x = 1/3. How do I take the domain into consideration?

Answer 9

well we have\[f'(x) =\frac{\pi}{2} \frac{1}{\cos^2(\frac{\pi x}{2})}\] right?

Answer 10

x=1/3 is in our domain and no problem with workin on that

Answer 11

\(f'(\frac{1}{3}) = \frac{\pi}{2cos^2(\frac{\pi}{6})}\)

Answer 12

well thats the tangent line slope

Answer 13

how can i find the other variables of y = mx+c?

Answer 14

the line is tangent to the \(f(x)\) so the point \((1/3,f(1/3))\) lies on the line also

Answer 15

line intersects \(f(x)\) at the point \(x=1/3\) a i right?

Answer 16

*am

Answer 17

I'm confused, how can I find the y intercept of f'(x) from here?

Answer 18

Just apply chain rule here
Use
\[\frac{ d }{ dx }(\tan(x))=\sec^2(x)\]
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\[\frac{d}{dx}\left(\tan \left(\frac{\pi x}{2}\right)\right)=\frac{1}{2} \pi \sec ^2\left(\frac{\pi x}{2}\right)\]

Answer 19

I can't use secant in this question though

Answer 20

You mean you want to find the angles?

Answer 21

just find the equation of the tangent

Answer 22

y=mx+c we know the value of m right till here?

Answer 23

yes, from the derivative

Answer 24

ok u just need to evalute c ... if we obtain a point of line we can find c right?