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

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