Question

find the equation of the tangent line to the curve ( y = 5xcos x ) at the point ( (pi,-5pi) ).
the equation of this tangent line can be written in the form ( y = mx + b ) where
( m=)
and ( b=)
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Explanation:

Step1: Find the derivative of the function

Use the product rule \((uv)^\prime = u^\prime v+uv^\prime\), where \(u = 5x\) and \(v=\cos x\).
\(u^\prime=5\), \(v^\prime=-\sin x\)
\(y^\prime=(5x\cos x)^\prime=5\cos x-5x\sin x\)

Step2: Calculate the slope \(m\)

Substitute \(x = \pi\) into \(y^\prime\)
\(m=y^\prime|_{x = \pi}=5\cos\pi-5\pi\sin\pi\)
Since \(\cos\pi=- 1\) and \(\sin\pi = 0\)
\(m=5\times(-1)-5\pi\times0=-5\)

Step3: Find the \(y -\)intercept \(b\)

Use the point - slope form \(y - y_1=m(x - x_1)\), where \((x_1,y_1)=(\pi,-5\pi)\) and \(m=-5\)
\(y+5\pi=-5(x - \pi)\)
Expand: \(y+5\pi=-5x + 5\pi\)
\(y=-5x\), so \(b = 0\)

Answer:

\(m=-5\), \(b = 0\)