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bearhunter [10]
3 years ago
8

Find dy/dx if y= (x²-3)³cos2x

Mathematics
1 answer:
Over [174]3 years ago
7 0

Answer:

dy/dx = 6x(x²-3)cos2x - 2(x²-3)³sin2x

Step-by-step explanation:

We use the Product Rule:

y= (x²-3)³cos2x

dy/dx =   (x²-3)³ d(cos2x)/dx + cos2x * d ((x²-3)³)/dx

=   (x²-3)³ * -2sin2x + cos2x * 3(x²-3)*2x

= -2sin2x*(x²-3)³ + 6xcos2x*(x²-3)

=  6x(x²-3)cos2x - 2(x²-3)³sin2x

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What is the difference between arcsin and sin^-1 ?
Len [333]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/2094473

_______________


Both refer to the inverse sine function.

The inverse sine (or arcsine) of x:

\mathsf{f(x)=arcsin(x)=sin^{-1}(x)}


where x is a real number in the domain of the function:

\mathsf{-1\le x\le 1}


and the arcsine function returns an angle in the interval  \mathsf{\left[-\,\frac{\pi}{2},\,\frac{\pi}{2}\right]:}

\mathsf{-\,\dfrac{\pi}{2}\le arcsin(x)\le \dfrac{\pi}{2}}\quad\longleftarrow\quad\textsf{range of the inverse sine function.}


So if you see anywhere one of these expressions below

\mathsf{\theta=arcsin(x)~~or~~\theta=sin^{-1}(x)}


then you should look for an angle \theta that satisfies the following conditions:

\footnotesize\begin{array}{l}\bullet\end{array}\normalsize\begin{array}{l}\mathsf{sin\,\theta=x;} \end{array}\\\\\\
\footnotesize\begin{array}{l}\bullet\end{array}\normalsize\begin{array}{l}\mathsf{\dfrac{\pi}{2} \le \theta\le\dfrac{\pi}{2}.} \end{array}


This angle \theta is called the inverse sine of the real number x.

______


Pay attention and do not mistake the arcsine function for the reciprocal of sine (which is cosecant); especially if you prefer or see that notation with an superscript -1. This one can be easily mistaken for an exponent:

\mathsf{sin^{-1}(x)=arcsin(x)\qquad\quad\checkmark}


but the reciprocal is something like

\mathsf{\big[sin(x)\big]^{-1}=\dfrac{1}{sin\,(x)}=csc(x)\qquad\quad(!!)}

and this last one has a total different meaning.


I hope this helps. =)

6 0
4 years ago
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