2022-03-20

Sinh3tcos2t

Vasquez

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

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

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

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

To apply the Chain Rule, set ${u}_{1}$ as $2t$.

$\mathrm{sinh}\left(3t\right)\left(\frac{d}{d{u}_{1}}\left[\mathrm{cos}\left({u}_{1}\right)\right]\frac{d}{dt}\left[2t\right]\right)+\mathrm{cos}\left(2t\right)\frac{d}{dt}\left[\mathrm{sinh}\left(3t\right)\right]$

The derivative of $\mathrm{cos}\left({u}_{1}\right)$ with respect to ${u}_{1}$ is $-\mathrm{sin}\left({u}_{1}\right)$.

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

Replace all occurrences of ${u}_{1}$ with $2t$.

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

Differentiate.

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

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

Multiply $2$ by $-1$.

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

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

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

Multiply $-2$ by $1$.

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

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

To apply the Chain Rule, set ${u}_{2}$ as $3t$.

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

The derivative of $\mathrm{sinh}\left({u}_{2}\right)$ with respect to ${u}_{2}$ is $\mathrm{cosh}\left({u}_{2}\right)$.

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

Replace all occurrences of ${u}_{2}$ with $3t$.

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

Differentiate.

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

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

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

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

Simplify the expression.

Multiply $3$ by $1$.

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

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

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

Reorder terms.

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

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