Question

problems with derivative

Answer 1

the correct method is
\[x=x_osin(\omega t) \\ \\ \frac{dx}{dt}=x_ocos(\omega t) \omega \\ \\ \frac{dx}{dt}=x_o \omega \cos(\omega t) \]
but why isn't
\[x=x_osin(\omega t) \\ \\ \frac{dx}{dt}=x_ocos(\omega t) t \\ \\ \frac{dx}{dt}=x_o t \cos(\omega t) \]
we're differentiating with respect to time right, isn't time get's differentiated?

Answer 2

just like what you do for normal equations
\[y=x^2 \\ \\ \frac{dy}{dx}=2x\]

Answer 3

by right it should be \[x=x_osin(\omega t) \\ \\ \frac{dx}{d \omega}=x_ocos(\omega t) \omega \\ \\ \frac{dx}{d \omega}=x_o \omega \cos(\omega t)\]

Answer 4

@ash2326

Answer 5

ur method is wrong!!

Answer 6

\[\frac{ d }{ dt }(xsin(yt))\]\[x*\frac{ d }{ dt }(\sin(yt))\]\[xcos(yt)*\frac{ d }{ dt }(yt)\]\[xycos(yt)\]

Answer 7

In,
$$y=x^2$$, we use the identity,
$$\frac{dax^n}{dx}=anx^{n-1}$$
but in,$$\frac{d\,[\omega_0\sin(\omega t)]}{dt}$$ we should use the chain rule.That is because we only know d(sin x)/dx and d(ax)/dx separtely but not d(sin(ax))/dx

take $$u=\omega t$$then, $$x=\omega_0\sin(u)$$
to find, dx/dt, we use the chain rule,

$$\frac{dx}{dt}=\frac{dx}{du}\frac{du}{dt}$$
$$\begin{align*}
\frac{dx}{dt}&=\frac{d\,[\omega_0\sin(u)]}{du}\frac{d[\omega t]}{dt}\\
&=\left(\omega_0\frac{d\,[\sin(u)]}{du}\right)\left(\omega\frac{ d[ t]}{dt}\right)\\
&=\omega_0\omega \frac{d\,\sin(u)}{du}\frac{dt}{dt}\\
&=\omega_0\omega[\cos(u)].[1]\\
&=\omega_0\omega\cos(\omega t)
\end{align*}$$

Answer 8

Looks like equations for Simple harmonic Motion?

Answer 9

@Calculator, it's because the derivative of the wt in cos(wt), with respect to time, is w, not t. Exactly like what you said with the x^2 example... like this

\[\frac{ d }{ dx } \sin(x^2) = \cos(x^2) \times (2x) = 2 x \cos (x^2) \]