find the derivative of $k(a)=sin^{3}acos^{4}a$ $k(a)=$

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u\cdot v$, then $y^\prime=u^\prime v + uv^\prime$. Let $u = \sin^{3}a$ and $v=\cos^{4}a$.

Step2: Find $u^\prime$

Use the chain - rule. If $u=\sin^{3}a$, let $t = \sin a$, then $u = t^{3}$. By the chain - rule, $\frac{du}{da}=\frac{du}{dt}\cdot\frac{dt}{da}$. $\frac{du}{dt}=3t^{2}=3\sin^{2}a$ and $\frac{dt}{da}=\cos a$, so $u^\prime = 3\sin^{2}a\cos a$.

Step3: Find $v^\prime$

Use the chain - rule. If $v=\cos^{4}a$, let $s=\cos a$, then $v = s^{4}$. By the chain - rule, $\frac{dv}{da}=\frac{dv}{ds}\cdot\frac{ds}{da}$. $\frac{dv}{ds}=4s^{3}=4\cos^{3}a$ and $\frac{ds}{da}=-\sin a$, so $v^\prime=-4\cos^{3}a\sin a$.

Step4: Calculate $k^\prime(a)$

Using the product - rule $k^\prime(a)=u^\prime v+uv^\prime$. [ \begin{align*} k^\prime(a)&=(3\sin^{2}a\cos a)\cos^{4}a+\sin^{3}a(-4\cos^{3}a\sin a)\ &=3\sin^{2}a\cos^{5}a - 4\sin^{4}a\cos^{3}a \end{align*} ]

Answer:

$3\sin^{2}a\cos^{5}a - 4\sin^{4}a\cos^{3}a$