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

\({dy\over dx}=xe^x({1\over x})+e^x.log \ x(1)+x \ log \ x(e^x)\)
\(=e^x+e^xlog \ x+xe^x log \ x=e^x(1+log \ x+xlog x).\)

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

Let y = x² (x + 1)³ (x + 2)⁴
Taking logarithm on both sides we have,
\(log\quad y=logx2+log(x+1)3+log(x+2)4=2logx+3log(x+1)+4log(x+2)\)
Differentiating both sides with respect to 'x' we have,
\(\frac { 1 }{ y } \frac { dy }{ dx } =\frac { 2 }{ x } +\frac { 3 }{ x+1 } +\frac { 4 }{ x+2 } \)
\(\Rightarrow \frac { dy }{ dx } =y\left[ \frac { 2 }{ x } +\frac { 3 }{ x+1 } +\frac { 4 }{ x+2 } \right] \Rightarrow \frac { dy }{ dx } ={ x }^{ 2 }{ (x+1) }^{ 3 }{ (x+2) }^{ 4 }\left[ \frac { 2 }{ x } +\frac { 3 }{ x+1 } +\frac { 4 }{ x+2 } \right] \)

Let y = \(\sqrt { { e }^{ \sqrt { x } } } \)
Differentiating both sides with respect to 'x' we have,
\(\frac { dy }{ dx } =\frac { d }{ dx } \sqrt { { e }^{ \sqrt { x } } } =\frac { 1 }{ 2\sqrt { { e }^{ \sqrt { x } } } } .\frac { d }{ dx } \left( { e }^{ \sqrt { x } } \right) =\frac { 1 }{ 2\sqrt { { e }^{ \sqrt { x } } } } .{ e }^{ \sqrt { x } }.\frac { d }{ dx } \left( \sqrt { x } \right) =\frac { 1 }{ 2\sqrt { { e }^{ \sqrt { x } } } } .{ e }^{ \sqrt { x } }.\frac { 1 }{ 2\sqrt { x } } \)
\(\frac { dy }{ dx } =\frac { { e }^{ \sqrt { x } } }{ 4\sqrt { x } .\sqrt { { e }^{ \sqrt { x } } } } \)

Given \({ x }^{ 4 }+{ x }^{ 2 }{ y }^{ 2 }+{ y }^{ 4 }=50\)  Differentiating both sides with respect to 'x' we have,
\(4{ x }^{ 3 }+{ x }^{ 2 }.2y\frac { dy }{ dx } +{ y }^{ 2 }.2x+4{ y }^{ 3 }\frac { dy }{ dx } =0\)
\(\Rightarrow \frac { dy }{ dx } (2{ x }^{ 2 }y+4{ y }^{ 3 })=-4{ x }^{ 3 }-2x{ y }^{ 2 }\)
\(\Rightarrow \frac { dy }{ dx } =-\frac { 4{ x }^{ 3 }-2x{ y }^{ 2 } }{ 2{ x }^{ 2 }y+4{ y }^{ 3 } } =-\frac { 2x(2{ x }^{ 2 }+{ y }^{ 2 }) }{ 2y({ x }^{ 2 }+2{ y }^{ 2 }) } \)
\( \therefore \frac { dy }{ dx } =-\frac { x(2{ x }^{ 2 }+{ y }^{ 2 }) }{ y({ x }^{ 2 }+2{ y }^{ 2 }) } \)

\( \sqrt{x y} =e^{(z-y)} \)
\((x y)^{1 / 2} =e^{x-y}\)
Take log on both sides
\(\frac{1}{2} \log x y=(x-y) \log e\)
\(\frac{1}{2} \log x y=x-y\)
\( \frac{1}{2} \cdot \frac{1}{x y}\left(x \frac{d y}{d x}+y\right) =1-\frac{d y}{d x}\)
\(\frac{1}{2 y} \frac{d y}{d x}+\frac{1}{2 x} =1-\frac{d y}{d x}\)
\(\frac{1}{2 y} \frac{d y}{d x}+\frac{d y}{d x} =1-\frac{1}{2 x}\)
\(\frac{d y}{d x}\left(\frac{1}{2 y}+1\right) =1-\frac{1}{2 x}\)
\(\frac{d y}{d x}=\frac{\frac{2 x-1}{2 x}}{\frac{1+2 y}{2 y}} =\frac{(2 x-1)}{2 x} \cdot \frac{2 y}{(1+2 y)}\)
\(=\frac{y(2 x-1)}{x(1+2 y)}\)

We have x = at² ; y = 2at
\({dy\over dx}={y'(t)\over x'(t)}={2a\over 2at}={1\over t}.\)

y = cos(tan x)
Take \(u=\tan x \Rightarrow \frac{d u}{d x}=\sec ^2 x\)
\(y=\cos u\)
\(\frac{d y}{d x}=\frac{d y}{d u} \cdot \frac{d u}{d x}=-\sin u \cdot\left(\sec ^2 x\right)\)
\(=-\sin (\tan x) \sec ^2 x\)

y = sin x + cos x
\(\frac{d y}{d x}=\cos x-\sin x\)