We have x = a(t - sin t), y = a(1 - cos t).
Now \({dx\over dt}=a(1-cos \ t);{dy\over dt}=a \ sin \ t\)
Therefore, \(\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{a \sin t}{a(1-\cos t)}=\frac{\sin t}{(1-\cos t)}\)

\(f(x)=\frac{x}{\sqrt{7-3 x}}\)
\(u=x \quad v=7-3 x\)
\(\frac{d u}{d x}=1 \quad \frac{d v}{d x}=-3\)
\(f(\dot{x})=u v^{\frac{-1}{2}}\)
\(\prime^{\prime}(x)=u \frac{d(v)^{-1 / 2}}{d x}+v^{-1 / 2} \frac{d u}{d x}\)
\(=u\left(-\frac{1}{2} v^{-1 / 2-1}\right) \frac{d v}{d x}+v^{-1 / 2}(1)\)
\(=u \frac{-1}{2} v^{-3 / 2}(-3)+v^{-1 / 2}\)
\(=\frac{+3 u}{2\left(v^{3 / 2}\right)}+v^{-1 / 2}\)
\(=\frac{+3 u+2 v}{2\left(v^{3 / 2}\right)}=\frac{+3 x+2(7-3 x)}{2(7-3 x)^{3 / 2}} \)
\(=\frac{3 x+14-6 x}{2(7-3 x) \sqrt{7-3 x}}=\frac{14-3 x}{2(7-3 x) \sqrt{7-3 x}}\)

y = sin x³
Differentiate with respect to x³
y^' = cos x²

X = a cos³ t; y = a sin³ t
\(\frac{d x}{d t}=\mathbf{3a} \cos ^2 t(-\sin t)\)
\(\frac{d y}{d t}=\mathbf{3a} \sin ^2 t(\cos t)\)
\(\frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{\operatorname{sen} \sin ^2+\cos t}{\operatorname{sacos}^2 t(-\sin t)}\)
\(=\frac{-\sin t}{\cos t}\)
\(=-\tan t\)

Let x = cos \(\theta\)
Then 4x³ − 3x = 4cos³ \(\theta\)- 3cos = cos3\(\theta\) and
f(x) = cos^-1(cos3\(\theta\) ) = 3\(\theta\) = 3cos^-1x
Therefore,f'(x) = 3\(({-1\over \sqrt{1-x^2}})={-3\over \sqrt{1-x^2}}.\)

\(y =\sin ^3 x+\cos ^3 x \)
\(u =\sin x \quad v=\cos x\)
\(\frac{d u}{d x} =\cos x \quad \frac{d v}{d x}=-\sin x\)
\(y =u^3+v^3\)
\(\frac{d y}{d x} =3 u^2 \frac{d u}{d x}+3 v^2 \frac{d v}{d x} \)
\(=3 \sin ^2 x(\cos x)+3 \cos ^2 x(-\sin x)\)
\(=3 \sin x \cos x[\sin x-\cos x]\)

\(y=5^{\frac{-1}{x}}\)
\(u =\frac{-1}{x}\)
\(\frac{d u}{d x} =1 / x^2\)
\(y =5^u\)
\(\frac{d y}{d x} =\frac{d y}{d u} \cdot \frac{d u}{d x}\)
\(=5^u \log 5\left(\frac{1}{x^2}\right)=\frac{5^{-1 / x} \log 5}{x^2} \) \({\left[\because d\left(a^x\right)=a^x \log a\right] }\)

We have, f(x) = x cos x.
Now f'(x) = - x sin x + cos x, and
f'(x) = - (x cos x + sin x) - sin x
= - x cos x - 2 sin x.

We have, y = x³ - 6x² - 5x + 3 and
y' = 3x² - 12x - 5 
y'' = 6x - 12
y''' = 6.

We have x = at² ; y = 2at
\({dy\over dx}={y'(t)\over x'(t)}={2a\over 2at}={1\over t}.\)