Prove a formula for the 2nth derivative of x sin (ax)

Define $f(x) = x \sin (ax)$ and prove the formula for the $2n$ th derivative of $f$ ,

$f^{(2n)}(x) = (-1)^n (a^{2n} x \sin (ax) - 2n a^{2n-1} \cos (ax)).$

Proof. The proof is by induction. For the case $n = 1$ we need to take two derivatives of $f$ (since $f^{(2n)} = f^{(2)} = f''$ ).

$\begin{align*} && f(x) &= x \sin (ax) \\ \implies && f'(x) &= \sin (ax) + ax \cos (ax) \\ \implies && f''(x) &= a \cos (ax) + a \cos (ax) - a^2 x \sin (ax) \\ &&&= (-1)^n(a^{2n} x \sin (ax) - 2n a^{2n-1} \cos (ax)). \end{align*}$

So, the formula holds for the case $n = 1$ . Assume then that the formula is true for some positive integer $k$ . Then the inductive hypothesis is that,

$f^{(2k)}(x) = (-1)^k \left( a^{2k} x \sin (ax) - 2k a^{2k-1} \cos (ax) \right).$

Now, we want to take two derivatives (to get a formula for $f^{(2(k+1))}(x)$ ).

$\begin{align*} f^{(2k+1)}(x) = \left(f^{(2k)}(x)\right)' &= \left( (-1)^k \left( a^{2k} x \sin (ax) - 2k a^{2k-1} \cos (ax)\right)\right)' \\[9pt] &= (-1)^k \big( a^{2k} \sin (ax) + a^{2k+1} x \cos (ax) + 2ka^{2k} \sin (ax) \big) \\[9pt] &= (-1)^k \big( (2k+1)a^{2k} \sin (ax) + a^{2k+1} x \cos (ax) \big). \end{align*}$

Now, we take another derivative,

$\begin{align*} f^{(2(k+1))}(x) &= \left( f^{(2k+1)}(x)\right)' \\[9pt] &= \left( (-1)^k \big( (2k+1)a^{2k} \sin (ax) + a^{2k+1} x \cos (ax) \big) \right)' \\[9pt] &= (-1)^k \big( (2k+1)a^{2k+1} \cos (ax) + a^{2k+1} \cos (ax) - a^{2k+2} x \sin (ax) \big) \\[9pt] &= (-1)^k \big( -a^{2(k+1)} x \sin (ax) + (2k+2)a^{2k+1} \cos (ax) \big) \\[9pt] &= (-1)^{k+1} \big(a^{2(k+1)} x \sin (ax) - 2(k+1)a^{2(k+1) - 1} \cos (ax) \big). \end{align*}$

Thus, the formula holds for $k+1$ . Hence, it holds for all positive integers $n. \qquad \blacksquare$

Stumbling Robot

Prove a formula for the 2nth derivative of x sin (ax)

Related

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment): Cancel reply