Find d y d x (a) y= x^ x^ x ...... (b) y= _e x+ _e x+ _e x+ (c) y… - Vidyadip

Question

Find $\frac{d y}{d x}$
(a) $y=\sin x^{\sin x^{\sin x ...... \infty}}$
(b) $y=\sqrt{\log _e x+\sqrt{\log _e x+\sqrt{\log _e x+\ldots \ldots \ldots \infty}}}$
(c) $y=e^{x+e^{x+e^{x+\ldots \infty}}}$

✓

Answer

(a) $y=\sin x^{\sin x^{\sin x}}$
or$
y=(\sin x)^y
$
taking logarithmic both sides
$
\begin{aligned}
\log _e y & =\log _e(\sin x)^y \\
\log _e y & =y \log _e(\sin x)
\end{aligned}
$
differentiate both sides w.r.t. $x$$
\begin{array}{l}
\quad \frac{1}{y} \frac{d y}{d x}=y \times \frac{1}{\sin x} \cdot \cos x+\log _e \sin x \cdot \frac{d y}{d x} \\
\Rightarrow\left(\frac{1}{y}-\log _e \sin x\right) \frac{d y}{d x}=y \cot x \\
\Rightarrow \quad \frac{d y}{d x}=\frac{y^2 \cot x}{1-y \log _e \sin x}
\end{array}
$
(b) $y=\sqrt{\log _c x+\sqrt{\log _c x+\sqrt{\log _c x+\ldots \ldots \infty}}}$
$\begin{aligned} \text { or } & y & =\sqrt{\log _c x+y} \\ \text { or } & y^2 & =\log _e
x+y\end{aligned}$
now differentiate w.r.t. $x$
$
\begin{aligned}
2 y \frac{d y}{d x} & =\frac{1}{x}+\frac{d y}{d x} \\
(2 y-1) \frac{d y}{d x} & =\frac{1}{x} \\
\Rightarrow \quad \frac{d y}{d x} & =\frac{1}{x(2 y-1)}
\end{aligned}
$
(c) $y=e^{x+e^{x+e^x.....\infty}}$
or $y=e^{x+y}$
taking logarithmic both sides
$\begin{aligned}
\therefore \quad \log _e y & =\log _e e^{x+y} \\
\log _c y & =(x+y) \log _e e
\end{aligned}$
or $\quad \log _c y=x+y \quad\left(\because \log _c e=1\right)$
differentiate w.r.t. $x$
$\frac{1}{y} \frac{d y}{d x}=1+\frac{d y}{d x}$
or $\quad\left(\frac{1}{y}-1\right) \frac{d y}{d x}=1$
$\frac{(1-y)}{y} \frac{d y}{d x}=1$
or $\quad \frac{d y}{d x}=\frac{y}{1-y}$

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