differentiate $g(t)=sin(3t + 5)cos(7t^{2}+9t)$. $g(t)=$

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u\cdot v$, then $y^\prime=u^\prime v + uv^\prime$. Let $u=\sin(3t + 5)$ and $v = \cos(7t^{2}+9t)$.

Step2: Differentiate $u$ with respect to $t$

Using the chain - rule, if $u=\sin(3t + 5)$, let $x = 3t+5$, then $\frac{du}{dt}=\frac{du}{dx}\cdot\frac{dx}{dt}$. Since $\frac{d}{dx}\sin(x)=\cos(x)$ and $\frac{d}{dt}(3t + 5)=3$, we have $u^\prime = 3\cos(3t + 5)$.

Step3: Differentiate $v$ with respect to $t$

Using the chain - rule, if $v=\cos(7t^{2}+9t)$, let $y = 7t^{2}+9t$, then $\frac{dv}{dt}=\frac{dv}{dy}\cdot\frac{dy}{dt}$. Since $\frac{d}{dy}\cos(y)=-\sin(y)$ and $\frac{d}{dt}(7t^{2}+9t)=14t + 9$, we have $v^\prime=-(14t + 9)\sin(7t^{2}+9t)$.

Step4: Apply the product - rule formula

$g^\prime(t)=u^\prime v+uv^\prime=3\cos(3t + 5)\cos(7t^{2}+9t)-\sin(3t + 5)(14t + 9)\sin(7t^{2}+9t)$.

Answer:

$3\cos(3t + 5)\cos(7t^{2}+9t)-(14t + 9)\sin(3t + 5)\sin(7t^{2}+9t)$