Question

Find an equation of the tangent line to the curve at the given point.
y = 8 x cos x

P = (pi , -8pi)

Answer 1

The tangent line at (pi,-8pi) is
y-(-8pi)=f'(pi)(x-pi)
y+8pi=f'(pi)(x-pi)

Answer 2

So find f'(x)
then find f'(x) evaluated at x=pi

Answer 3

how do i find f'(x)??? @freckles

Answer 4

by differentiating f(x)=8xcos(x) w.r.t x

Answer 5

You would find product rule useful.

Answer 6

8x-sinx @freckles

Answer 7

(uv)'=u'*v+v'*u

Answer 8

\[\frac{d}{dx}(uv)=u \frac{d}{dx}v+v \frac{d}{dx}u \\ \frac{d}{dx}8xcos(x)=\frac{d}{dx}(8x)(\cos(x)) \\ \frac{d}{dx}(8x)(\cos(x))=(8x) \frac{d}{dx} (\cos(x))+\cos(x) \frac{d}{dx}(8x)\]

Answer 9

for the derivative i got 8(cos(x)-xsin(x)), is this correct? @freckles

Answer 10

\[f'(x)=8x(-\sin(x))+\cos(x)(8) \\= 8[-x \sin(x)+\cos(x)] \\=8[\cos(x)-x \sin(x)\]
looks good