If y = x , find d y d x - Vidyadip

Question

If $y =\sqrt{\tan \sqrt{x}}$, find $\frac{ d y}{ d x}$

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Answer

$y =\sqrt{\tan \sqrt{x}}$
Differentiating w.r.t. $x$, we get
$ \frac{ d y}{ d x}=\frac{ d }{ d x}(\sqrt{\tan \sqrt{x}})$
$=\frac{1}{2 \sqrt{\tan \sqrt{x}}} \cdot \frac{ d }{ d x}(\tan \sqrt{x})$
$=\frac{1}{2 \sqrt{\tan \sqrt{x}}} \cdot \sec ^2(\sqrt{x}) \cdot \frac{ d }{ d x}(\sqrt{x})$
$=\frac{\sec ^2 \sqrt{x}}{2 \sqrt{\tan \sqrt{x}}} \cdot \frac{1}{2 \sqrt{x}}$
$\therefore \frac{ d y}{ d x}=\frac{\sec ^2 \sqrt{x}}{4 \sqrt{x \sqrt{\tan \sqrt{x}}}} $

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"Let $f(x)=x^2+5$ and $g(x)=e^x+3$ then
$f[g(x)]=$ and $g[f(x)]=$
Now $f^{\prime}(x)=$ and $g^{\prime}(x)=$
The derivative off $[g(x)]$ w.r. t. $x$ in terms of $f$ and $g$ is
Therefore $\frac{d}{d x}[f[g(x)]]=$ and $\left[\frac{d}{d x}[f[g(x)]]\right]_x=0=$
The derivative of $g[f(x)]$ w.r. t. $x$ in terms of $f$ and $g$ is
Therefore $\frac{d}{d x}[ g [ f ( x )]]=\ldots . . . .$. and $\left[\frac{d}{d x}[ g [ f ( x )]]\right]_{ x =1}=$
Hint basket: $\left\{f^{\prime}[g(x)] \cdot g^{\prime}(x), 2 e^{2 x}+6 e^x, 8, g^{\prime}[f(x)] \cdot f^{\prime}(x), 2 x e^{x 2+5},-2 e^6, e^{2 x}+6 e^x+14, e^{x 2}+\right.$ $\left.5+3,2 x, e^x\right\}$