Chapter 4

$$ y=\left(x^{2}+1\right)^{\sin x} $$

$$ y=\sqrt[3]{\frac{x\left(x^{2}+1\right)}{\left(x^{2}-1\right)^{2}}} $$

$$ x^{3}+x^{2} y+y^{2}=0 $$

$$ x^{2}+y^{2}-4 x-10 y+4=0 $$

$$ x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}} $$

$$ \tan ^{-1} y-y+x=0 $$

$$ e^{x}-e^{y}=y-x $$

$$ x+y=e^{x-y} $$

$$ x^{3}-2 x^{2} y^{2}+5 x+y-5=0 $$

$$ x+\sqrt{x y}+y=a $$

$$ \tan ^{-1}\left(\frac{y}{x}\right)=\ln \sqrt{x^{2}+y^{2}} $$

$$ e^{x} \sin y-e^{-y} \cos x=0 $$

$$ e^{y}+x y=e $$

$$ x^{2}+5 x y+y^{2}-2 x+y-6=0 $$

$$ \text { If } 3 \sin (x y)+4 \cos (x y)=5, \text { then show that } \frac{d y}{d x}=-\frac{y}{x} \text { . } $$

$$ \text { If } y=\sqrt{\sin x+y}, \text { then find }\left(\frac{d y}{d x}\right)_{x=0 \atop y=1} \text { . } $$

$$ \text { If } \sin y=x \sin (a+y), \text { then find }\left(\frac{d y}{d x}\right)_{x=y=0} $$

$$ x=a(t-\sin t), y=a(1-\cos t) $$

$$ x=k \sin t-\sin k t, y=k \cos t+\cos k t $$

$$ x=2 \ln \cot t, y=\tan t+\cot t $$

$$ x=e^{c t}, y=e^{-c t} $$

$$ \left\\{\begin{array}{l} x=a \cos ^{3} t \\ y=b \sin ^{3} t \end{array}\right. $$

$$ \left\\{\begin{array}{l} x=t^{3}+3 t+1 \\ y=t^{3}-3 t+1 \end{array}\right. $$

$$ \left\\{\begin{array}{l} x=a(\cos t+t \sin t) \\ y=a(\sin t-t \cos t) \end{array}\right. $$

$$ \left\\{\begin{array}{l} x=e^{t} \cos t \\ y=e^{t} \sin t \end{array}\right. $$

$$ x=e^{-t}, \quad y=t^{3} $$

$$ x=\sec t, \quad y=\tan t $$

$$ x=\frac{a \sin t}{1+b \cos t}, \quad y=\frac{c \cos t}{1+b \cos t} $$

$$ x=\ln \left(1+t^{2}\right), y=t-\tan ^{-1} t $$

$$ x=t^{2}+2, y=\frac{t^{3}}{3}-t $$

$$ x=e^{-t^{2}}, y=\tan ^{-1}(2 t+1) $$

$$ x=4 \tan ^{2} \frac{t}{2}, y=a \sin t+b \cos t $$

$$ \text { Given } x=\sin ^{-1}\left(t^{2}-1\right), y=\cos ^{-1} 2 t, \text { find }\left(\frac{d y}{d x}\right)_{t=\frac{1}{4}} $$

$$ \text { Given } x=\sin ^{-1} t, y=\sqrt{1-t^{2}}, \text { find }\left(\frac{d y}{d x}\right)_{t=\frac{1}{2}} $$

$$ \text { Differentiate } \tan ^{-1} \frac{\sqrt{1+x^{2}}-1}{x} \text { w.r.t. } \tan ^{-1} x $$

$$ \text { Differentiate } \sin ^{-1} \frac{1-x}{1+x} \text { w.r.t. } \sqrt{x} $$

$$ \text { Differentiate } \sin ^{-1} \frac{1-x}{1+x} \text { w.r.t. } \sqrt{x} $$

$$ \text { Differentiate } x^{\sin ^{-1} x} \text { w.r.t. } \sin ^{-1} x $$

$$ \text { Differentiate } \sec ^{-1} \frac{1}{2 x^{2}-1} \text { w.r.t. } \sqrt{1-x^{2}} $$

$$ \text { Differentiate } \frac{\tan ^{-1} x}{1+\tan ^{-1} x} \text { w.r.t. } \tan ^{-1} x \text { . } $$

$$ \text { 3. Let } U=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) \text { and } V=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right), \text { then find } \frac{d U}{d V} . \text { Ans. } $$

$$ \text { Let } f(x)=\frac{x^{2}}{1-x^{2}}, x \neq 0, \pm 1, \text { then find } f^{\prime}(2) \text { . } $$

$$ \text { If } f(x)=(x+1) \tan ^{-1}\left(e^{-2 x}\right), \text { then find } f^{\prime}(0) \text { . } $$

$$ \text { If } f(x)=\log _{x^{2}}(\ln x), \text { then find } f^{\prime}(e) \text { . } $$

$$ \text { Find the derivative of the function } \cot ^{-1} \sqrt{\cos 2 x} \text { at } x=\frac{\pi}{6} \text { . } $$

$$ \text { If } f(x)=\tan ^{-1} \sqrt{\frac{1+\sin x}{1-\sin x}}, 0 \leq x \leq \frac{\pi}{2}, \text { then find } f^{\prime}\left(\frac{\pi}{6}\right) \text { . } $$

$$ \text { Find the derivative w.r.t. } x \text { of the function }\left(\log _{\cos x} \sin x\right)\left(\log _{\sin x} \cos x\right)^{-1}+\sin ^{-1} \frac{2 x}{1+x^{2}} \text { at } x=\frac{\pi}{4} $$

$$ \text { Given } f(x)=|x-1|+|x-3|, \text { find } f^{\prime}(2) \text { . } $$

$$ f(x)=\ln x \cdot \cos ^{-1} x, \text { find } f^{\prime}(1) $$

$$ f(x)=\left(x^{2}-1\right) \cos ^{-1} x \text { , find } f^{\prime}(-1) \& f^{\prime}(1) $$