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I need to find the equation of the tangent line to $f′(x) = 4 \sin x + 3 \cos x$ at $x= π/3$. I'm trying to incorporate the slope point formula.

Progress

This is what I got: $f'(x)= 4 \cos x- 3 \sin x$ at $x= π/3$, $$f(π/3)= 4 \cos π/3 - 3 \sin π/3 = 4(1/2)-3(\sqrt{3}/2) = ?? $$

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    $\begingroup$ If you want the tangent line to $f'(x)$ at $x$ then take the derivative of $f'(x)$. The slope is $f''(\frac{\pi}{3})$ $\endgroup$
    – recmath
    Sep 21, 2014 at 17:45
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    $\begingroup$ It's the straight line with slope $f'(\pi / 3)$ that goes through $(\pi/3, f(\pi/3)$ - just work those quantities out and plug it all into the formula for a straight line. $\endgroup$
    – Matt Rigby
    Sep 21, 2014 at 17:45
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    $\begingroup$ Welcome to Math SE ! People are ready to help you but show your efforts, describe what you tried and tell where you are stuck. Cheers :-) $\endgroup$ Sep 21, 2014 at 17:46
  • $\begingroup$ tangent line to $f$ or to $f'$? $\endgroup$
    – ir7
    Sep 21, 2014 at 17:46
  • $\begingroup$ it is tangent line to f $\endgroup$ Sep 21, 2014 at 17:48

1 Answer 1

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you will get $f''(x)=4\cos(x)-3\sin(x)$ and $f''(\pi/3)=2\sqrt{3}+\frac{3}{2}$ further we have $f'(\pi/3)=\frac{3}{2}+2\sqrt{3}$ and you can calculate the tangent line $y=mx+n$ Sonnhard.

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