3 ; y' (x0)-?...

July 2022 1 13 Report

y= 7sin3x+2cos5x, x0=п/3 ; y' (x0)-?

Answers & Comments

bearcab

Verified answer

$y'_x=7*3*\cos(3x)-2*5\sin(5x)$

$y'_x(x_0)=7*3*\cos\left(3*\frac{\pi}{3}\right)-2*5\sin\left(5*\frac{\pi}{3}\right)$

$3*7*\cos\left(3*\frac{\pi}{3}\right)-2*5\sin\left(5*\frac{\pi}{3}\right)=3*7*\cos\pi-10*\sin\left(\pi+\frac{2\pi}{3}\right))$

$3*7*\cos\pi-10*\sin\left(\pi+\frac{2\pi}{3}\right)=$

$-21-10*(\sin\pi\cos\frac{2\pi}{3}+\cos\pi\sin\frac{2\pi}{3})=-21-10*(-\sin\frac{2\pi}{3})$

$-21+10*\frac{\sqrt{3}}{2}=-21+5\sqrt{3}$

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