Single Variable Calculus, Chapter 3, 3.5, Section 3.5, Problem 85

a.) Show that $\displaystyle \frac{d}{dx} (\sin ^n x \cos nx) = n \sin^{n - 1} x \cos (n + 1) x$

Suppose $n$ is an integer.

Using Chain Rule,


$
\begin{equation}
\begin{aligned}

\frac{d}{dx} (\sin^n x \cos n x) =& \frac{d}{dx} (\sin^n x) \cdot \cos n x + \sin^n x \cdot \frac{d}{dx} (\cos n x)
\\
\\
\frac{d}{dx} (\sin^n x \cos n x) =& n \sin ^{n - 1} x \cos x \cdot \cos n x + \sin^n x \cdot (- \sin (nx)) (n)
\\
\\
\frac{d}{dx} (\sin^n x \cos n x) =& n \sin ^{n-1} x (\cos x \cos (nx) - \sin x \sin (nx))

& \text{Using the sum of the angles for cosine..}
\\
\\
& \cos (A + B) = \cos A \cos B - \sin A \sin B
\\
\\
\frac{d}{dx} (\sin^n x \cos n x) =& n \sin^{n - 1} x (\cos (x + nx))
\\
\\
\frac{d}{dx} (\sin^n x \cos n x) =& n \sin^{n - 1} x (\cos x (1 + n))
\\
\\
\frac{d}{dx} (\sin^n x \cos n x) =& n \sin^{n - 1} x [\cos (n + 1)x]
\end{aligned}
\end{equation}
$




b.) Find a formula for the derivative of $y =\cos^n x$ that is similar to the one in part (a).

Using Chain Rule,


$
\begin{equation}
\begin{aligned}

\frac{d}{dx} (\cos^n x \cos nx) =& \frac{d}{dx} (\cos ^n x) \cdot \cos n x + \cos ^n x \cdot \frac{d}{dx} (\cos n x)
\\
\\
\frac{d}{dx} (\cos^n x \cos nx) =& n \cos ^{n - 1} x (-\sin x) \cdot \cos n z + \cos^n x \cdot (- \sin nx)(n)
\\
\\
\frac{d}{dx} (\cos^n x \cos nx) =& -n \cos^{n - 1} x (\sin x \cos (nx) + \cos x \sin (nx))
\\
\\
& \text{Using the sum of angles for sine}
\\
\\
& \sin(A + B) = \sin A \cos B + \cos A \sin B
\\
\\
\frac{d}{dx} (\cos^n x \cos nx) =& - n \cos^{n - 1} x [\sin (x + n x)]
\\
\\
\frac{d}{dx} (\cos^n x \cos nx) =& -n \cos^{n - 1} x [\sin (1 + n) x]
\end{aligned}
\end{equation}
$