We have: \(log{ x }^{ y }=log\quad { e }^{ x-y }\)
\(ylogx=(x-y)\)
\(y(1+log\quad x)=x\)
\(y=\frac { x }{ 1+log\quad x } \)
\(\frac { dy }{ dx } =\frac { \left( 1+log\quad x \right) .1-x\left( 0+\frac { 1 }{ x } \right) }{ { \left( 1+logx \right) }^{ 2 } } \)
\(=\frac { 1+log\quad x-1 }{ { \left( 1+logx \right) }^{ 2 } } =\frac { log\quad x }{ { \left( 1+logx \right) }^{ 2 } } \)
\(\frac { dy }{ dx } =\frac { log\quad x }{ { \left( log\quad e+log\quad x \right) }^{ 2 } } \)

We have: \(y={ sin }^{ -1 }\left[ \frac { 6x-4\sqrt { 1-4{ x }^{ 2 } } }{ 5 } \right] \)
\(={ sin }^{ -1 }\left( \frac { 6x }{ 5 } -\frac { 4 }{ 5 } \sqrt { 1-4{ x }^{ 2 } } \right) \)
\(={ sin }^{ -1 }\left( (2x)\frac { 3 }{ 5 } -\frac { 4 }{ 5 } \sqrt { 1-4{ x }^{ 2 } } \right) \)
\(={ sin }^{ -1 }\left( (2x)\sqrt { 1-{ \left( \frac { 4 }{ 5 } \right) }^{ 2 } } -\left( \frac { 4 }{ 5 } \right) \sqrt { 1-{ \left( 2x \right) }^{ 2 } } \right) \)
\(={ sin }^{ -1 }\left( 2x \right) -{ sin }^{ -1 }\frac { 4 }{ 5 } \)
Hence, \(\frac { dy }{ dx } =\frac { 1 }{ \sqrt { 1-{ \left( 4x \right) }^{ 2 } } } .(2)-0=\frac { 2 }{ \sqrt { 1-4{ x }^{ 2 } } } \)

One way is to solve for y and rewrite the above as
\(y=x-\pi\)
\(\text {But then }\frac{d y}{d x}=1\)
Alternatively, directly differentiating the relationship w.r.t., x, we have
\(\frac{d}{d x}(x-y)=\frac{d \pi}{d x}\)
Recall that \(\frac{d \pi}{d x}\) means to differentiate the constant function taking value p everywhere w.r.t., x. Thus
\(\frac{d}{d x}(x)-\frac{d}{d x}(y)=0\)
which implies that
\(\frac{d y}{d x}=\frac{d x}{d x}=1\)

\(x^y+y^x=1\)
Let \(u=x^y, v=y^x\)
Hence,
u+v=1
Differentiating both sides w.r.t. x.
\(\frac{d(v+u)}{d x}=\frac{d(1)}{d x} \)
\(\frac{d v}{d x}+\frac{d u}{d x}=0\)
(Derivative of constant is 0 )

Let y=tan(2x+3)=tan t, where t=2x+3
\(\frac { dy }{ dt } ={ sec }^{ 2 }\ t\ and\ \)
\(\frac { dt }{ dx } =2(1)+0=2.\)
\(By\quad chain\quad rule,\frac { dy }{ dx } =\frac { dy }{ dt } .\frac { dt }{ dx } \)
\(={ sec }^{ 2 }\ t.2\)
\(=2 { sec }^{ 2 }(2x+3)\)

\(We\quad have:f(x)={ x }^{ 3 }+{ x }^{ 2 }-1\)
Which is polynomial function
\(and\ { D }_{ f }=R\)
\(Let\quad c\in { D }_{ f }\)
\(Then\ \lim _{ x\rightarrow c }{ f(x) } =\lim _{ x\rightarrow c }{ ({ x }^{ 3 }+{ x }^{ 2 }-1) } \)
\(={ c }^{ 3 }+{ c }^{ 2 }-1=f(c)\)
\(\Rightarrow \) f is continuous at x = c.
But c is arbitrary.
Hence, f is continuous at each of its domains.

The function is clearly defined at every point and f(c) = c for every real number c. Also
\(\lim _{ x\rightarrow c }{ f(x) } =\lim _{ x\rightarrow c }{ x } =c.\)
\( \lim _{ x\rightarrow c }{ f(x) } =f(c) \) and hence the function is continuous at every real number.
Having defined continuity of a function at a given point, now we make a natural extension of this definition to discuss continuity of a function