We know that, domain set is (0, \(\infty\)), range set is R and the curve passing the point (1, 0)
Case (i) when a > 1 and x > 0
\((f(x)=\log _a x= \begin{cases} < 0 & \text { if } 0 < x <1 \\ =0 & \text { if } x=1 \\ > 0 & \text { if } x > 1\end{cases} \)
We noticed that as x increases, f(x) is also increases.
\(\therefore \) The graph is as shown in the figure.

Case (ii) when 0 < a < 1 and x > 0
\((f(x)=\log _a x= \begin{cases} > 0 & \text { if } 0 < x < 1 \\ =0 & \text { if } x=1 \\ < 0 & \text { if } x > 1\end{cases} \)
We noticed that x increases, the value of f(x) decreases.
\(\therefore \) The graph is as shown in the figure

\(\underset { x\rightarrow 0 }{ lim } \frac { 2sinx-sin2x }{ { x }^{ 3 } } \)= \(\underset { x\rightarrow 0 }{ lim } \frac { 2sinx-2sinx\quad cosx }{ { x }^{ 3 } } \)
= \(\underset { x\rightarrow 0 }{ lim } \frac { 2sinx(1-cosx) }{ { x }^{ 3 } } =2.\underset { x\rightarrow 0 }{ lim } \frac { sinx }{ x } \underset { x\rightarrow 0 }{ lim } \frac { { 2sin }^{ 2 }\frac { x }{ 2 } }{ { x }^{ 2 } } \)  \(\left[ \therefore 1-cosx=2{ sin }^{ 2 }\frac { x }{ 2 } \right] \) 
= \(4(1).\underset { x\rightarrow 0 }{ lim } \frac { { sin }^{ 2 }\frac { x }{ 2 } }{ \left( \frac { x }{ 2 } \right) ^{ 2 } \times \ \ 4 } \left[ \therefore \underset { \phi \rightarrow 0 }{ lim } \quad \frac { sin\phi }{ \phi } =1 \right] \) [Multiplying and dividing by 4in the denominator]
= \(\frac { 4 }{ 4 } .\underset { x\rightarrow 0 }{ lim } \left( \frac { sin\frac { 2 }{ x } }{ \frac { 2 }{ x } } \right) \)
=1 x 1 = 1

\(\underset { x\rightarrow -3 }{ lim } \frac { { x }^{ 3 }+27 }{ { x }^{ 5 }+243 } =\underset { x\rightarrow -3 }{ lim } \frac { { x }^{ 3 }-\left( { -3 } \right) ^{ 3 } }{ { x }^{ 5 }-\left( { -3 } \right) ^{ 5 } } \)
Dividing the numerator and denominator by (x -3)
= \(\frac { \underset { x\rightarrow -3 }{ lim } \frac { { x }^{ 3 }-({ 3 })^{ 3 } }{ x-3 } }{ \frac { { x }^{ 5 }-\left( 3 \right) ^{ 5 } }{ x-3 } } =\frac { \underset { x\rightarrow -3 }{ lim } \frac { { x }^{ 3 }\left( { -3 } \right) ^{ 3 } }{ x-3 } }{ \underset { x\rightarrow -3 }{ lim } \frac { { x }^{ 5 }-\left( { -3 } \right) ^{ 5 } }{ x-3 } } \)
\(=\frac { 3\left( -3 \right) ^{ 3-1 } }{ 5\left( -3 \right) ^{ 5-1 } } \quad \left[ \therefore \underset { x\rightarrow a }{ lim } \frac { { x }^{ n }-{ a }^{ n } }{ x-a } ={ n }a^{ n-1 } \right] \)
= \(\frac { 3\left( -3 \right) ^{ 2 } }{ 5\left( -3 \right) ^{ 4 } } =\frac { 3(9) }{ 5(81) } =\frac { 3 }{ 5\left( 9 \right) } =\frac { 1 }{ 15 } \)

Given e^y(x+1)=1    ....(1)
Differentiating with respect to 'x' we get,
\({ e }^{ y }(1)+(x+1){ e }^{ y }\frac { dy }{ dx } =0\)   [product rule]
\(\Rightarrow { e }^{ y }+(1)\frac { dy }{ dx } =0\quad [using\quad (1)]\)
\(\Rightarrow \frac { dy }{ dx } =-{ e }^{ y }\)...(2)
Differentiating again with respect to 'x' we get,
\(\frac { d }{ dx } \left( \frac { dy }{ dx } \right) =\frac { d }{ dx } \left( -{ e }^{ y } \right) \)
\(\Rightarrow \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =-{ e }^{ y }.\frac { dy }{ dx } \)
\(=\left( \frac { dy }{ dx } \right) \left( \frac { dy }{ dx } \right) \) [using (2)]
\(={ \left( \frac { dy }{ dx } \right) }^{ 2 }\)
\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } ={ \left( \frac { dy }{ dx } \right) }^{ 2 }\)
Hence proved.

If x = 0, y = 1. The curve cuts the y axis at (0,1)
The curve will not meet the x axis for all real values of x
 

u = sin³x ; v = cos³x
\(\frac { du }{ dx } =3sin^{ 2 }x; \frac { d }{ dx } (sinx)=3{ sin }^{ 2 }xcosx\)
\(\therefore \frac { du }{ dv } =\frac { \frac { du }{ dx } }{ \frac { dv }{ dx } } =\frac { 3{ sin }^{ 2 }x\cos x }{ -3{ cos }^{ 2 }x\sin x } =-\frac { \sin x }{ \cos x } =-\tan x\)

\(f(x)=\frac{x-1}{3 x+1} \)
\(f\left(\frac{1}{x}\right)=\frac{\frac{1}{x}-1}{\frac{3}{x}+1}=\frac{1-x}{3+x} \)
\(\frac{1}{f(x)}=\frac{3 x+1}{x-1} \)
 

f(x) = \(\frac { x+1 }{ x-1 } \)
\(\text { LHS }=f(f(x)) =f\left(\frac{x+1}{x-1}\right)=\frac{\frac{x+1}{x-1}+1}{\frac{x+1}{x-1}-1} \)
\(=\frac{x+1+x-1}{x+1-x+1}=\frac{2 x}{2}=x=R H S \)

Given x² +xy +y² = 100
Differentiating with respect to 'x' we get
2x +x.\(\frac { dy }{ dx } \) + y(1) + 2y \(\frac { dy }{ dx } \) = 0
\(\Rightarrow\) \(\frac { dy }{ dx } \) (x+2y) = (x +2y) = -2 x - y
\(\Rightarrow\)  \(\frac { dy }{ dx } \) = \(\frac { -(2x+y) }{ x+2y } \)

Let f(x) = \({ x }^{ \frac { 2 }{ 3 } }\)
f(x+h) = (x+h)\(^{ \frac { 2 }{ 3 } }\)
\(\frac { d }{ dx } (f(x))=\underset { h\rightarrow 0 }{ lim } \frac { f(x+h)-f(x) }{ h } \)
= \(\underset { h\rightarrow 0 }{ lim } \frac { (x+h)^{ \frac { 2 }{ 3 } }-x^{ \frac { 2 }{ 3 } } }{ x+h-x } \) [adding and subtracting x in the denominator]
= \(\frac { 2 }{ 3 } .x^{ \frac { 2 }{ 3 } -1 }\)
\(\left[ \therefore \underset { x\rightarrow a }{ Lt } \frac { { x }^{ n }-{ a }^{ n } }{ x-a } =n-a^{ n-1 } \right] =\frac { 2 }{ 3 } { x }^{ \frac { -1 }{ 3 } }\)
\(\therefore \frac { d }{ dx } \left( x^{ \frac { 2 }{ 3 } } \right) -\frac { 2 }{ 3 } .x^{ \frac { -1 }{ 3 } }\)

x = a (θ - sin θ),y = a(1- cosθ)
\(\frac { dx }{ d\theta } =a(1-cos\theta );\ \frac { dy }{ d\theta } =a(sin\theta )\)
\(={ 2asin }^{ 2 }\frac { \theta }{ 2 } ;\) \(=2asin\frac { \theta }{ 2 } cos\frac { \theta }{ 2 } \)
\( \frac { dy }{ dx } =\frac { \frac { dy }{ d\theta } }{ \frac { dx }{ d\theta } } =\frac { 2asin\frac { \theta }{ 2 } cos\frac { \theta }{ 2 } }{ 2a\ sin^{ 2 }\frac { \theta }{ 2 } } =\frac { cos\frac { \theta }{ 2 } }{ sin\frac { \theta }{ 2 } } =cot\left( \frac { \theta }{ 2 } \right) \)

\(y_1=\frac{1}{x}, y_2=\frac{-1}{x^2}\)

\(f(x)=\frac{x^2+1}{x} \)
\(f\left(\frac{1}{x}\right)=\frac{\frac{1}{x^2}+1}{\frac{1}{x}}=\frac{1+x^2}{x^2} \times \frac{x}{1}=\frac{1+x^2}{x}=f(x) \)

Let f(x)=x³-27
\(\therefore \frac { d }{ dx } (f(x))=\begin{matrix} \underset { h\rightarrow 0 }{ lim } & \frac { f(x+h)-f(x) }{ h } \end{matrix}\)
\(\frac { d }{ dx } (f(x))=\begin{matrix} \underset { h\rightarrow 0 }{ lim } & \frac { \left[ { \left( x+h \right) }^{ 3 }-27 \right] -({ x }^{ 3 }-27) }{ h } \end{matrix}\)
\(=\begin{matrix} \underset { h\rightarrow 0 }{ lim } & \frac { { x }^{ 3 }+3{ x }^{ 2 }h+3x{ h }^{ 2 }+{ h }^{ 3 }-27-{ x }^{ 3 }+27 }{ h } \end{matrix}\)
\(=\begin{matrix} \underset { h\rightarrow 0 }{ lim } & \frac { 3{ x }^{ 2 }h+3x{ h }^{ 2 }+{ h }^{ 3 } }{ h } \end{matrix}\)
\(\begin{matrix} \underset { h\rightarrow 0 }{ lim } & \frac { h(3{ x }^{ 2 }+3x{ h }+{ h }^{ 2 }) }{ h } \end{matrix}\)
\(=\begin{matrix} \underset { h\rightarrow 0 }{ lim } & 3{ x }^{ 2 }+3x{ h }+{ h }^{ 2 }\end{matrix}\)
\(=3{ x }^{ 2 }+3x{ (0) }+0)\)
\(=3{ x }^{ 2 }\)
\(\therefore \frac { d }{ dx } ({ x }^{ 3 }-27)=3{ x }^{ 2 }\)

Let \(y=\frac { sinx+cosx }{ sinx-cosx } \)
Differentiating with respect to 'x' we get,
\(\frac { dy }{ dx } =\frac { (sinx-cosx).\frac { d }{ dx } (sinx+cosx)-(sinx+cosx).\frac { d }{ dx } (sinx-cosx) }{ { (sinx-cosx) }^{ 2 } } \)
\(=\frac { (sinx-cosx)(cosx-sinx)-(sinx+cosx)(cosx+sinx) }{ { (sinx-cosx) }^{ 2 } } \)
\(=\frac { (sinxcosx-{ sin }^{ 2 }x-cos^{ 2 }x+sinx.cosx)-({ sin }^{ 2 }x+cos^{ 2 }x+2sinxcosx) }{ { (sinx-cosx) }^{ 2 } } \)
\(=\frac { 2sinxcox-1-1-2sinxcox }{ { (sinx-cosx) }^{ 2 } } \)
\(=\frac { -2 }{ { (sinx-cosx) }^{ 2 } } \left[ \therefore sin^{ 2 }x+cos^{ 2 }x=1 \right] \)

Let f(x) = log_e (x + 1)
f(x + h) = log_e (x + h + 1)
\(\therefore\) \(\frac { d }{ dx } (f(x))=\lim _{ h\rightarrow 0 }{ \frac { f(x+h)-f(x) }{ h } } \)
= \(\lim _{ h\rightarrow 0 }{ \frac { { log }_{ e }(x+h+1)-{ log }_{ e }(x+1) }{ h } } =\lim _{ h\rightarrow 0 }{ \frac { { log }_{ e }\left( \frac { x+h+1 }{ x+1 } \right) }{ h } } \)
\(=\lim _{h \rightarrow 0} \frac{\log \left(1+\frac{h}{x+1}\right)}{h}\)
=  \(\lim _{ h\rightarrow 0 }{ \frac { { log }_{ e }\left( 1+\frac { h }{ x+1 } \right) }{ \frac { h }{ x+1 } \times x+1 } } \)  
= \(\frac { 1 }{ x+1 } \left[ \lim _{ \frac { h }{ x+1 } \rightarrow 0 }{ \frac { { log }_{ e }\left( 1+\frac { h }{ x+1 } \right) }{ \frac { h }{ x+1 } } } \right] \) 
\(\text {Let z}=\frac{h}{x+1} \text{ As h }\rightarrow 0 z \rightarrow 0\)
\(=\frac{1}{x+1} \lim _{z \rightarrow 0} \frac{\log (1+z)}{z}=\frac{1}{x+1}(1)=\frac{1}{x+1} \)
\(\frac{d}{d x}(\log (x+1))=\frac{1}{x+1}\)