We have, y = a^x +x^a+x^x+a^a
Let v = x^x
log v = x log x
\(\frac { 1 }{ v } \frac { dv }{ dx } =x+\frac { 1 }{ x } +logx\)
\(\frac { dv }{ dx } =v\left| 1+logx \right| \)
= x^x(1+lodx)
\(\frac { dy }{ dx } ={ a }^{ x }loga+a{ x }^{ a-1 }+\frac { dv }{ dx } +0\)
\(\Rightarrow \frac { dy }{ dx } ={ a }^{ x }loga+a{ x }^{ a-1 }+{ x }^{ x }(1+logx)\)

We have, y = tan^-1\(\sqrt { \frac { 1-x }{ 1+x } } \)
Put x = cos2\(\theta\)
\(\Rightarrow\)2\(\theta\) = cos^-1x
\(\Rightarrow\)\(\theta\) = 1/2cos^-1x
y = tan^-1\(\sqrt { \frac { 1-cos2\theta }{ 1+cos2\theta } } \)
y = \({ tan }^{ -1 }\sqrt { \frac { 2{ sin }^{ 2 }\theta }{ 2{ cos }^{ 2 }\theta } } \)
(\(\because\)cos2\(\theta\) = 2cos²\(\theta\)-1 = 1-2sin²\(\theta\)
y = tan^-1(tan \(\theta\))
y = \(\theta\) = 1/2cos^-1x
\(\frac { dy }{ dx } =\frac { 1 }{ 2 } \left( \frac { -1 }{ \sqrt { 1-{ x }^{ 2 } } } \right) =\frac { -1 }{ 2\sqrt { 1-{ x }^{ 2 } } } \)​​​​​​​

We have y = \(f({ e }^{ { { sin }^{ -1 } } }2x)\)
dy/dx = f^'\(({ e }^{ { { sin }^{ -1 } } }2x)\) x d/dx \(({ e }^{ { { sin }^{ -1 } } }2x)\)
= f^'\(({ e }^{ { { sin }^{ -1 } } }2x)\)x\(({ e }^{ { { sin }^{ -1 } } }2x)\)xd/dx(sin^-12x)
= f^'\(({ e }^{ { { sin }^{ -1 } } }2x)\)x\(({ e }^{ { { sin }^{ -1 } } }2x)\)x\(\frac { 1 }{ \sqrt { 1-4{ x }^{ 2 } } } \times 2\)
= \(\frac { 2{ e }^{ sin-1 }2x }{ \sqrt { 1-4{ x }^{ 2 } } } { f }^{ ' }({ e }^{ sin-1 }2x)\)

y'=0

Absolute value function f(x) = Ix - 1I is continuous at x = 1 but not differentiable at x = 1.

f( -1 ) = 1 + 5 = 6,
\(\lim _{x \rightarrow-1} f(x)=\lim _{x \rightarrow-1}\left(x^{2}+5\right)=1+5=6 \)
\(\text { As } \lim _{x \rightarrow-1} f(x)=f(-1)\)
Hence, continuous at x = -1.

\(Let\quad y={ x }^{ sinx\quad }...(1)\)
\(Taking\quad logs.,\quad log\quad y={ log\quad x }^{ sinx }=sinxlogx.\)
\(Diff.\quad w.r.t.\quad x,\quad \frac { 1 }{ y } .\frac { dy }{ dx } =sinx.\frac { 1 }{ x } +logx.cosx\)
\(\frac { dy }{ dx } =y\left[ \frac { sinx }{ x } +cosxlogx \right] \)
\(={ x }^{ sinx\quad }\left[ \frac { sinx }{ x } +cosxlogx \right] \)

Define g by g(x) = 1 – x + |x| and h by h(x) = |x| for all real x.
Then (h o g) (x) = h(g (x)) = h (1– x + | x |)
= |1– x + | x || = f(x)
we have seen that h is a continuous function.
Hence g being a sum of a polynomial function and the modulus function is continuous.
But then f being a composite of two continuous functions is continuous.

The function is defined at x = 0 and its value at x = 0 is 1. When x \(\ne\) 0, the function is given by a polynomial. Hence
\(\lim _{ x\rightarrow 0 }{ f(x) } =\lim _{ x\rightarrow 0 }{ { x }^{ 3 }+3 } =0+3=3\)
Since the limit of f at x = 0 does not coincide with f(0), the function is not continuous at x = 0. It may be noted that x = 0 is the only point of discontinuity for this function.

Let \(y=(\sin x)^x+\sin ^{-1} \sqrt{x}\)
Let \(u=(\sin x)^x \ v=\sin ^{-1} \sqrt{x}\)
y=u+v
Differentiating both sides w.r.t. x.
\( \frac{d y}{d x}=\frac{d(u+v)}{d x} \)
\( \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\)

Putting x = \(cos2\theta \left\{ 2tan^{ -1 }\sqrt { \frac { 1-x }{ 1+x } } \right\} \) we get 
\(2tan^{ -1 }\sqrt { \frac { 1-cos2\theta }{ 1+cos2\theta } } \)
i,e.,\(2tan^{ -1 }\sqrt { \frac { 2sin^{ 2 }\theta }{ 2cos^{ 2 }\theta } } \)
\(=2\quad tan^{ -1 }\left( tan\theta \right) \)
\(=2\theta =cos^{ -1 }x\)
Hence \(y=e^{ sin^{ 2 } }xcos^{ -1 }x\)
\(\Rightarrow logy=sin^{ 2 }x+log\left( cos^{ 1 }x \right) \)
\(\Rightarrow \frac { 1 }{ y } \times \frac { dy }{ dx } =2sinxcosx+\frac { 1 }{ cos^{ -1 }x } \times \frac { -1 }{ \sqrt { 1-x^{ 2 } } } \)
= sin 2x \(-\frac { 1 }{ cos^{ 1 }x\sqrt { 1-x^{ 2 } } } \)
\(\Rightarrow \frac { dy }{ dx } =e^{ sin^{ 2 } }xcos^{ -1 }x\left[ sin2x-\frac { 1 }{ cos^{ -1 }x\sqrt { 1-x^{ 2 } } } \right] \)