\({dy\over dx}=3x^2+10x+3.\)

Given f(x) = x |x| = \(\begin{matrix} { x }^{ 2 }, & x\ge 0 \\ -x^{ 2 } & x<0 \end{matrix}\)
f'(0^-) = \(\underset { x\rightarrow 0^{ - } }{ lim } \frac { f(x)-f(0) }{ x-0 } =\underset { x\rightarrow 0^{ - } }{ lim } \frac { -x^{ 2 }-0 }{ x-0 } =\underset { x\rightarrow 0^{ - } }{ lim } \frac { -{ x }^{ 2 } }{ x } =\underset { x\rightarrow 0^{ - } }{ lim } (-x)=0\)  ....(1)
∴ f'(0⁺) = \(\underset { x\rightarrow 0^{ + } }{ lim } \frac { f(x)-f(0) }{ x-0 } =\underset { x\rightarrow 0^{ + } }{ lim } \frac { -x^{ 2 }-0 }{ x-0 } =\underset { x\rightarrow 0^{ - } }{ lim } \frac { -{ x }^{ 2 } }{ x } =\underset { x\rightarrow 0^{ + } }{ lim } (x)=0\) .....(2)
\(\therefore\) f'(0^-) = f'(0⁺) = 0
\(\therefore\) It is differentiable at x = 0.

\(f(t)=\sqrt[3]{1+\tan t} \)
\(Take u=1+\tan t\)
\( \frac{d u}{d t} =\sec ^2 t\)
\(f(t) =u^{1 / 3}\)
\(f^{\prime}(t) =\frac{d f}{d u} \times \frac{d u}{d t}=\frac{1}{3} u^{-2 / 3}\left(\sec ^2 t\right)\)
\(=\frac{1}{3}(1+\tan t)^{-2 / 3}\left(\sec ^2 t\right)\)

If y = tan^-1\(({1+x\over 1-x})\)
Let \({1+x\over 1-x}=t\)
Then, y = tan^-1t
\({dy\over dx}={d\over dt}(tan^{-1}t).{dt\over dx}\)
\(={1\over 1+t^2}.{(1-x).1-(1+x)(-1)\over (1-x)^2}\)
\(=\frac{1}{1+\left(\frac{1+x}{1-x}\right)^2} \cdot \frac{(1-x)+(1+x)}{(1-x)^2}=\frac{1}{1+x^2}\)

y = e^x sin^x
\(\frac{d y}{d x}=\frac{d}{d x}\left(e^x\right) \sin x_1+e^x \frac{d}{d x}(\sin x)\)
\(=e^x \sin x+e^x \cos x\)
\(=e^x(\sin x+\cos x)\)

\( \sin y =x \sin (a+y) \)
\( x =\frac{\sin y}{\sin (a+y)} \)
\(\frac{d x}{d y} =\frac{\sin (a+y) \cos y-\sin y \cos (a+y)}{\sin ^2(a+y)} \)
\( =\frac{\sin (a+y-y)}{\sin ^2(a+y)} \)
[Since sin(A - B) = sin A cos B - cos A sin B]
\(\frac{d x}{d y}=\frac{\sin a}{\sin ^2(a+y)}\)
Take reciprocal.
\(\frac{d y}{d x}=\frac{\sin ^2(a+y)}{\sin a}\) \(a \neq n \pi\)
Hence proved.

cos(xy) = x
Differentiate w.r. to x.
\( -\sin (x y)\left[1 \cdot y+x \cdot \frac{d y}{d x}\right] =1\)
\(-y \sin (x y)-x \sin (x y) \frac{d y}{d x} =1 \)
\(x \sin (x y) \frac{d y}{d x} =-y \sin x y-1\)
\(\therefore \frac{d y}{d x} =\frac{-(1+y \sin x y)}{x \sin (x y)}\)
Hence Proved.

\(s(t)=\sqrt[4]{\frac{t^3+1}{t^3-1}}\)
u = t³ + 1 and v = t³ - 1
\(\frac{d u}{d t} =3 t^2 \quad \frac{d v}{d t}=3 t^2\)
\(s(t) =u^{1 / 4} v^{-1 / 4}\)
\(s^{\prime}(t) =u^{1 / 4} \frac{d}{d t}\left(v^{-1 / 4}\right)+v^{-1 / 4} \frac{d}{d t}\left(u^{1 / 4}\right) \)
\(=u^{1 / 4}\left[\frac{-1}{4} v^{-1 / 4} \frac{d v}{d t}\right]+v^{-1 / 4}\left[\frac{1}{4} u^{1 / 4} \cdot \frac{d u}{d t}\right]\)
\(=\frac{-1}{4} u^{1 / 4} v^{-5 / 4}\left(3 t^2\right)+\frac{v^{-1 / 4}}{4} u^{-3 / 4}\left(3 t^2\right) \)
\(=\frac{1}{4}\left(3 t^2\right) u^{-3 / 4} v^{-5 / 4}\left[-u^1+v^1\right] \)
\(=\frac{3 t^2}{4 u^{3 / 4} v^{5 / 4}}[-u+v] \)
\(=\frac{3 t^2}{4\left(t^3+1\right)^{3 / 4}\left(t^3-1\right)^{5 / 4}}\left[-t^3-1+t^3-1\right] \)
\(=\frac{3 t^2}{4\left(t^3+1\right)^{3 / 4}\left(t^3-1\right)^{5 / 4}}(-2) \)
\(=\frac{-3 t^2}{2\left(t^3+1\right)^{3 / 4}\left(t^3-1\right)^{5 / 4}}\)

\(f^{\prime}\left(2^{-}\right)=\lim _{x \rightarrow 2^{-}} \frac{f(x)-f(2)}{x-2}=\lim _{x \rightarrow 2^{-}} \frac{x+1-(2+1)}{x-2}\)
\(=\lim _{x \rightarrow 2^{-}} \frac{x+1-3}{x-2}=\lim _{x \rightarrow 2^{-}} \frac{x-2}{x-2}=1\)
\(f^{\prime}\left(2^{+}\right)=\lim _{x \rightarrow 2^{+}} \frac{f(x)-f(2)}{x-2}=\lim _{x \rightarrow 2^{+}} \frac{(2 x-1)-(4-1)}{x-2}\)
\(=\lim _{x \rightarrow 2^{+}} \frac{2 x-1-3}{x-2}=\lim _{x \rightarrow 2^{+}} \frac{2 x-4}{x-2}\)
\(=\lim _{x \rightarrow 2^{+}} \frac{2(x-2)}{(x-2)}=2\)
\(f^{\prime}\left(2^{-}\right) \neq f^{\prime}\left(2^{+}\right)\)
\(\therefore f^{\prime}(2) \text { does not exist. }\)

\( p v=81\)
\( p=\frac{81}{v}=\frac{81}{9}=9 \)
Diff w. r. to v
\( p(1)+v \cdot \frac{d p}{d v} =0\)
\(v \frac{d p}{d v} =-p \)
\(\frac{d p}{d v} =\frac{-p}{v}=\frac{-9}{9}=-1\)
\(\frac{d p}{d v} =-1\)

\(\frac{d x}{d t} =\frac{\left(1+t^{2}\right)(-2 t)-\left(1-t^{2}\right)(2 t)}{\left(1+t^{2}\right)^{2}} \)
\(=\frac{-2 t-2 t^{3}-2 t+2 t^{3}}{\left(1+t^{2}\right)^{2}}=\frac{-4 t}{\left(1+t^{2}\right)^{2}} \)
\(\frac{d y}{d t} =\frac{\left(1+t^{2}\right)(2)-2 t(2 t)}{\left(1+t^{2}\right)^{2}}=\frac{2+2 t^{2}-4 t^{2}}{\left(1+t^{2}\right)^{2}} \)
\(=\frac{2-2 t^{2}}{\left(1+t^{2}\right)^{2}}=\frac{2\left(1-t^{2}\right)}{\left(1+t^{2}\right)^{2}} \)
\(\frac{d y}{d x} =\frac{d y / d t}{d x / d t}=\frac{2^{\left(1-t^{2}\right)} /\left(1+t^{2}\right)^{2}}{-4 t /\left(1+t^{2}\right)^{2}} \)
\(=\frac{1-t^{2}}{-2 t}=-\frac{x}{y} \)

\(y =m x+c \)
\(f(x) =m x+c \Rightarrow f(0)=c=1 \)
\(\therefore c =1 \)
\(f^{\prime}(x) =m \)
\(f^{\prime}(0) =1=m \)
\(\therefore m =1 \)
\(\therefore f(x) =x+1 \)
\(f(2) =2+1=3 \)

\(y=f\left(x^{2}+2\right) \)
\(\frac{d y}{d x} =f^{\prime}\left(x^{2}+2\right)(2 x) \)
\(\text { At } x =1, \frac{d y}{d x}=f^{\prime}(1+2)(2)=f^{\prime}(3)(2) \)
\(=5(2)=10 \)