Take u = g(x) = x² + 1 and f(u) =\(\sqrt{u}\)
\(\therefore F(x)=(fog)(x)=f(g(x))\)
Since \(f'(u)={1\over2}u^{1\over2}={1\over 2\sqrt{u}}\) and
\(g'(x)=2x,\) we get
\(F'(x)=f'(g(x))g'(x)\)
\(={1\over 2\sqrt{x^2+1}}.2x={x\over \sqrt{x^2+1}}\).

g(t) = t³ cos t
\(g^{\prime}(t)=\frac{d}{d t}\left(t^3\right) \cos t+t^3 \frac{d}{d t}(\cos t)\)
\(=3 t^2 \cos t-t^3 \sin t\)

\(|x|=\left\{\begin{array}{ccc} -x & \text { if } & x<0 \\ x & \text { if } & x>0 \end{array}\right.\)
\(\sin |x|=\left\{\begin{array}{lll} -\sin x & \text { if } & x<0 \\ \sin x & \text { if } & x>0 \end{array}\right.\)
\(f^{\prime}\left(0^{-}\right)=\lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}=\lim _{x \rightarrow 0^{-}} \frac{-\sin x-0}{x}\)
\(=\lim _{x \rightarrow 0^{-}} \frac{-\sin x}{x}\)
= -1 \(\left[\text { Since } \lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1\right]\)
\(f^{\prime}\left(0^{+}\right)=\lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0}=\lim _{x \rightarrow 0^{+}} \frac{\sin x-0}{x-0}\)
\(=\lim _{x \rightarrow 0^{+}} \frac{\sin x}{x}=1\)

\(\therefore f^{\prime}\left(0^{-}\right) \neq f^{\prime}\left(0^{+}\right)\)
\(\therefore\) It is not differentiable at x = 0

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.

\(f(x)=-x^2+2\)
\(f(x+h)=-(x+h)^2+2=-x^2-h^2-2 x h+2\)
\(f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\)
\(=\lim _{h \rightarrow 0} \frac{-x^2-h^2-2 x h+2+x^2-2}{h}\)
\(=\lim _{h \rightarrow 0} \frac{+h(-h-2 x)}{h}\)
= -0 - 2x
\(f^{\prime}(x)=-2 x\)

The greatest integer function \(f(x)=\lfloor x\rfloor\) is not continuous at every integer point n, since \(\left.\lim _{x \rightarrow n^{-}} \mid x\right\rfloor=n-1\) and \(\lim _{x \rightarrow n^{+}}\lfloor x\rfloor=n .\) Thus f'(n) does not exist.

Step (i) f(5) = - 5(5)² + 7 × 5 = - 125 + 35 = - 90.
For any \(\triangle x\neq 0,\)
f(5 + \(\triangle\)x) = - 5(5 + \(\triangle\)x)² + 7(5 + \(\triangle\)x) = - 90 - 43\(\triangle\)x - 5(\(\triangle\)x)².
Step (ii) \(\triangle\)y = f(5 + \(\triangle\)x) - f(5)
= - 90 - 43\(\triangle\)x - 5(\(\triangle\)x)² + 90 = - 43\(\triangle\)x - 5(\(\triangle\)x)²
= \(\triangle\)x [- 43 - 5\(\triangle\)x].
Step (iii) \({\triangle \ y\over \triangle x}=-43-5\triangle x\)
Step (iv) m_tan = \(lim_{\triangle x \rightarrow0}{\triangle y\over \triangle x}\)= - 43.

\(y=\sqrt{1+2 \tan x}\)
\(u =1+2 \tan x \)
\(\frac{d u}{d x} =2 \sec ^2 \cdot x \)
\(y =\sqrt{u}=u^{1 / 2}\)
\(\frac{d y}{d x} =\frac{d y}{d u} \cdot \frac{d u}{d x}=1 / 2 u^{1 / 2-1}\left(2 \sec ^2 x\right) \)
\(=1 / 2 u^{-1 / 2}\left(2 \sec ^2 x\right)=\frac{1}{2 \sqrt{u}}\left(2 \sec ^2 x\right)\)
\(=\frac{\sec ^2 x}{\sqrt{1+2 \tan x}}\)

\(y=e^{-m x}\)
Take \(u=-m x\)
\(\frac{d u}{d x} =-m(1)=-m \)
\(y =e^u \)
\(\frac{d y}{d x} =\frac{d y}{d u} \cdot \frac{d u}{d x}=e^u(-m)=e^{-m x}(-m) \)
\(=-m e^{-m x}=-m y\)

\(F(x)=\left(x^3+4 x\right)^7\)
Take \(u=x^3+4 x \Rightarrow \frac{d u}{d x}=3 x^2+4\)
\(F(x)=\dot{u}^7\)
\(F^{\prime}(x)=\frac{d F}{d u} \cdot \frac{d u}{d x}=7 u^6\left(3 x^2+4\right)\)
\(=7\left(x^3+4 x\right)^6\left(3 x^2+4\right)\)

\(y=e^{-x} \cdot \log x\)
\(\frac{d y}{d x}=e^{-x} \frac{d}{d x}(\log x)+\log x \frac{d}{d x}\left(e^{-x}\right)\)
\(=e^{-x}\left(\frac{1}{x}\right)+\log x\left(-e^{\top x}\right) \quad\left[\therefore d\left(e^{n x}\right)=n e^{n x}\right]\)
\(=e^{-x}\left[\frac{1}{x}-\log x\right]\)

\(y=\frac{\tan x-1}{\sec X}\)
\(y=\frac{\frac{\sin x}{\cos x}-1}{1 / \cos x}=\frac{\sin x-\cos x}{\cos x} \cdot \frac{\cos x}{1}\)
\(y=\sin x-\cos x\)
\(\frac{d y}{d x}=\cos x+\sin x\)

\(\frac{d y}{d x}=e^x+\cos x\)

Given f is differentiable
\(\therefore f^{\prime}\left(1^{-}\right)=f^{\prime}\left(1^{+}\right)=1 \)
\(f^{\prime}\left(1^{-}\right) =\lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x+1}=\lim _{x \rightarrow 1^{-}} \frac{\left(a x^{2}-b\right)-(a-b)}{x+1} \)
\(=\lim _{x \rightarrow 1^{-}} \frac{a x^{2}-b-a+b}{x+1}=\lim _{x \rightarrow 1^{-}} \frac{a\left(x^{2}-1\right)}{x+1} \)
\(=\lim _{x \rightarrow 1^{-}}-a(x+1) \)
\(=a(1+1)=2 a \)
\(\therefore f^{\prime}\left(1^{+}\right) =\lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1}=\lim _{x \rightarrow 1^{+}} \frac{\frac{1}{x}-1}{x-1} \)
\(=\lim _{x \rightarrow 1^{+}} \frac{1-x}{x(x-1)}=\lim _{x \rightarrow 1^{+}} \frac{-1}{x}=-1\)
\(\therefore 2 a =-1 \)
\(a =\frac{-1}{2} \)
\(\text { and } f(1)=1\)
\(a-b=1 \)
\(-1 / 2-1=b \)
\(b=-3 / 2 \)

\(\text { Given } f(0)=5\)
\(\lim _{x \rightarrow 0} \frac{f(x)-5}{x} =4 \)
\(\lim _{x \rightarrow 0} \frac{f(x)-f(0)}{x-0} =4 \)
\(f^{\prime}(0) =4 \)
\(g(x) =\left(x^{2}+2 x+1\right) f(x) \)
\(g^{\prime}(x) =\left(x^{2}+2 x+1\right) f^{\prime}(x)+(2 x+2) f(x) \)
\(g^{\prime}(0) =(1) f^{\prime}(0)+2 f(0) \)
\(=1(4)+2(5) \)
\(=4+10=14 \)

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

\(y=\frac{(1-x)^{2}}{x^{2}}=\frac{1+x^{2}-2 x}{x^{2}}=x^{-2}+1-\frac{2}{x} \)
\(\frac{d y}{d x}=-2 x^{-3}+\frac{2}{x^{2}}=\frac{-2}{x^{3}}+\frac{2}{x^{2}} \)

\( y=\log _{10} x=\frac{1}{\log _x 10}\)
Diff  w. r. to \(\log _x 10\)
\(y^{\prime}=\frac{-1}{\left(\log _x 10\right)^2}=-\left(\log _{10} x\right)^2 \)

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

\(\text { Given } e^{x+5 \log x}=e^{x} \cdot e^{5 \log x}=e^{x} \cdot e^{\log x^{5}}=x^{5} e^{x}\)
\(\frac{d}{d x}\left(e^{x+5 \log x}\right) =\frac{d}{d x}\left(x^{5} e^{x}\right)=e^{x} x^{5}+e^{x}\left(5 x^{4}\right) \)
\(=e^{x} x^{4}(x+5) \)

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

\(\text { Given } y=\frac{1}{a-z}\)
\(a-z =\frac{1}{y} \)
\(a-\frac{1}{y} =z \)
\(z =a-\frac{1}{y} \)
\(\frac{d z}{d y}=0+\frac{1}{y^{2}}=\frac{1}{y^{2}} =\frac{1}{1 /(a-z)^{2}} \)
\(=(a-z)^{2} \)