Question

$$V_{1}=5\frac{d}{dt}(3\ \cos\ 15ot)+1\frac{d}{dt}(4\ sin|5ot)|$$

Answer

$$V=15*d^2*t^2*cos(15*o)+20*d^2*t^2*b*s*o*sin(a)$$

Solution


Remove parentheses.
\[V\times 1=5ddt\times 3(\cos{15o})t+1\times ddt\times 4(\sin{a})bs\times 5ot\]
Simplify  \(V\times 1\)  to  \(V\).
\[V=5ddt\times 3(\cos{15o})t+1\times ddt\times 4(\sin{a})bs\times 5ot\]
Take out the constants.
\[V=(5\times 3)ddtt\cos{15o}+1\times ddt\times 4(\sin{a})bs\times 5ot\]
Simplify  \(5\times 3\)  to  \(15\).
\[V=15ddtt\cos{15o}+1\times ddt\times 4(\sin{a})bs\times 5ot\]
Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[V=15{d}^{2}{t}^{2}\cos{15o}+1\times ddt\times 4(\sin{a})bs\times 5ot\]
Take out the constants.
\[V=15{d}^{2}{t}^{2}\cos{15o}+(4\times 5)ddttbso\sin{a}\]
Simplify  \(4\times 5\)  to  \(20\).
\[V=15{d}^{2}{t}^{2}\cos{15o}+20ddttbso\sin{a}\]
Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[V=15{d}^{2}{t}^{2}\cos{15o}+20{d}^{2}{t}^{2}bso\sin{a}\]