Question

find the first derivative of $z = \sin(\sin^{7}(t))$.
$\frac{dz}{dt}=$

Explanation:

Step1: Apply chain - rule

Let $u = \sin^{7}(t)$. Then $z=\sin(u)$. The chain - rule states that $\frac{dz}{dt}=\frac{dz}{du}\cdot\frac{du}{dt}$. First, find $\frac{dz}{du}$. Since $z = \sin(u)$, $\frac{dz}{du}=\cos(u)$.

Step2: Find $\frac{du}{dt}$

Let $v=\sin(t)$. Then $u = v^{7}$. Using the chain - rule again, $\frac{du}{dt}=\frac{du}{dv}\cdot\frac{dv}{dt}$. Since $\frac{du}{dv}=7v^{6}$ and $\frac{dv}{dt}=\cos(t)$, substituting $v = \sin(t)$ back in, we get $\frac{du}{dt}=7\sin^{6}(t)\cos(t)$.

Step3: Calculate $\frac{dz}{dt}$

Substitute $u=\sin^{7}(t)$ into $\frac{dz}{du}$ and use $\frac{du}{dt}=7\sin^{6}(t)\cos(t)$ in the chain - rule formula $\frac{dz}{dt}=\frac{dz}{du}\cdot\frac{du}{dt}$. So $\frac{dz}{dt}=\cos(\sin^{7}(t))\cdot7\sin^{6}(t)\cos(t)=7\cos(\sin^{7}(t))\sin^{6}(t)\cos(t)$.

Answer:

$7\cos(\sin^{7}(t))\sin^{6}(t)\cos(t)$