Sinh3tcos2t    

${\displaystyle \sinh \left(3t\right)\cos \left(2t\right)}$

Differentiate using the Product Rule which states that ${\displaystyle \frac{d}{dt}\left[f\left(t\right)g\left(t\right)\right]}$ is ${\displaystyle f\left(t\right)\frac{d}{dt}\left[g\left(t\right)\right]+g\left(t\right)\frac{d}{dt}\left[f\left(t\right)\right]}$ where ${\displaystyle f\left(t\right)=\sinh \left(3t\right)}$ and ${\displaystyle g\left(t\right)=\cos \left(2t\right)}$.

${\displaystyle \sinh \left(3t\right)\frac{d}{dt}\left[\cos \left(2t\right)\right]+\cos \left(2t\right)\frac{d}{dt}\left[\sinh \left(3t\right)\right]}$

Differentiate using the chain rule, which states that ${\displaystyle \frac{d}{dt}\left[f\left(g\left(t\right)\right)\right]}$ is ${\displaystyle f\prime \left(g\left(t\right)\right)g\prime \left(t\right)}$ where ${\displaystyle f\left(t\right)=\cos \left(t\right)}$ and ${\displaystyle g\left(t\right)=2t}$.

To apply the Chain Rule, set ${\displaystyle u_1}$ as ${\displaystyle 2t}$.

${\displaystyle \sinh \left(3t\right)\left(\frac{d}{d{u}_1}\left[\cos \left(u_1\right)\right]\frac{d}{dt}\left[2t\right]\right)+\cos \left(2t\right)\frac{d}{dt}\left[\sinh \left(3t\right)\right]}$

The derivative of ${\displaystyle \cos \left(u_1\right)}$ with respect to ${\displaystyle u_1}$ is ${\displaystyle -\sin \left(u_1\right)}$.

${\displaystyle \sinh \left(3t\right)(-\sin \left(u_1\right)\frac{d}{dt}\left[2t\right])+\cos \left(2t\right)\frac{d}{dt}\left[\sinh \left(3t\right)\right]}$

Replace all occurrences of ${\displaystyle u_1}$ with ${\displaystyle 2t}$.

${\displaystyle \sinh \left(3t\right)(-\sin \left(2t\right)\frac{d}{dt}\left[2t\right])+\cos \left(2t\right)\frac{d}{dt}\left[\sinh \left(3t\right)\right]}$

Differentiate.

Since ${\displaystyle 2}$ is constant with respect to ${\displaystyle t}$, the derivative of ${\displaystyle 2t}$ with respect to ${\displaystyle t}$ is ${\displaystyle 2\frac{d}{dt}\left[t\right]}$.

${\displaystyle \sinh \left(3t\right)(-\sin \left(2t\right)\left(2\frac{d}{dt}\left[t\right]\right))+\cos \left(2t\right)\frac{d}{dt}\left[\sinh \left(3t\right)\right]}$

Multiply ${\displaystyle 2}$ by ${\displaystyle -1}$.

${\displaystyle \sinh \left(3t\right)(-2\sin \left(2t\right)\frac{d}{dt}\left[t\right])+\cos \left(2t\right)\frac{d}{dt}\left[\sinh \left(3t\right)\right]}$

Differentiate using the Power Rule which states that ${\displaystyle \frac{d}{dt}\left[t^n\right]}$ is ${\displaystyle n{t}^{n-1}}$ where ${\displaystyle n=1}$.

${\displaystyle \sinh \left(3t\right)(-2\sin \left(2t\right)\cdot 1)+\cos \left(2t\right)\frac{d}{dt}\left[\sinh \left(3t\right)\right]}$

Multiply ${\displaystyle -2}$ by ${\displaystyle 1}$.

${\displaystyle \sinh \left(3t\right)(-2\sin \left(2t\right))+\cos \left(2t\right)\frac{d}{dt}\left[\sinh \left(3t\right)\right]}$

Differentiate using the chain rule, which states that ${\displaystyle \frac{d}{dt}\left[f\left(g\left(t\right)\right)\right]}$ is ${\displaystyle f\prime \left(g\left(t\right)\right)g\prime \left(t\right)}$ where f(t)=sinh(t)${\displaystyle f\left(t\right)=\sinh \left(t\right)}$ and ${\displaystyle g\left(t\right)=3t}$.

To apply the Chain Rule, set ${\displaystyle u_2}$ as ${\displaystyle 3t}$.

${\displaystyle \sinh \left(3t\right)(-2\sin \left(2t\right))+\cos \left(2t\right)\left(\frac{d}{d{u}_2}\left[\sinh \left(u_2\right)\right]\frac{d}{dt}\left[3t\right]\right)}$

The derivative of ${\displaystyle \sinh \left(u_2\right)}$ with respect to ${\displaystyle u_2}$ is ${\displaystyle \cosh \left(u_2\right)}$.

${\displaystyle \sinh \left(3t\right)(-2\sin \left(2t\right))+\cos \left(2t\right)\left(\cosh \left(u_2\right)\frac{d}{dt}\left[3t\right]\right)}$

Replace all occurrences of ${\displaystyle u_2}$ with ${\displaystyle 3t}$.

${\displaystyle \sinh \left(3t\right)(-2\sin \left(2t\right))+\cos \left(2t\right)\left(\cosh \left(3t\right)\frac{d}{dt}\left[3t\right]\right)}$

Differentiate.

Since ${\displaystyle 3}$ is constant with respect to ${\displaystyle t}$, the derivative of ${\displaystyle 3t}$ with respect to ${\displaystyle t}$ is ${\displaystyle 3\frac{d}{dt}\left[t\right]}$.

${\displaystyle \sinh \left(3t\right)(-2\sin \left(2t\right))+\cos \left(2t\right)\left(\cosh \left(3t\right)\left(3\frac{d}{dt}\left[t\right]\right)\right)}$

Differentiate using the Power Rule which states that ${\displaystyle \frac{d}{dt}\left[t^n\right]}$ is ${\displaystyle n{t}^{n-1}}$ where ${\displaystyle n=1}$.

${\displaystyle \sinh \left(3t\right)(-2\sin \left(2t\right))+\cos \left(2t\right)\left(\cosh \left(3t\right)(3\cdot 1)\right)}$

Simplify the expression.

Multiply ${\displaystyle 3}$ by ${\displaystyle 1}$.

${\displaystyle \sinh \left(3t\right)(-2\sin \left(2t\right))+\cos \left(2t\right)(\cosh \left(3t\right)\cdot 3)}$

Move ${\displaystyle 3}$ to the left of ${\displaystyle \cosh \left(3t\right)}$.

${\displaystyle \sinh \left(3t\right)(-2\sin \left(2t\right))+\cos \left(2t\right)(3\cdot \cosh \left(3t\right))}$

Reorder terms.

$-2\sin \left(2t\right)\sinh \left(3t\right)+3\cos \left(2t\right)\cosh \left(3t\right)$