Question

Solve and write the answer. Please do not expect me to mark you brainiest if you just write
I don't know the answer​

Answer 1

$\bigstar$ Given :

• $\bf y=x^{x^x^{......\infty}}$

$\bigstar$ To Prove :

• $\bf \dfrac{dy}{dx}=\dfrac{y^2}{x(1-ylogx)}$

$\bigstar$ Proof :

$\bf Let\ y=x^y$

Apply log on both sides , we get ,

$\implies \bf log\ y=log\ (x^y)\\\\\implies \bf log\ y=y.log\ x$

Now , Differentiate with respect to x on both sides , we get ,

$\implies \bf \dfrac{d}{dx}(log\ y)=\dfrac{d}{dx}(y.log\ x)\\\\\implies \bf \dfrac{1}{y}.\dfrac{dy}{dx}=log\ x.\dfrac{dy}{dx}+y.\dfrac{1}{x}\\\\\implies \bf \dfrac{dy}{dx}(\dfrac{1}{y}-log\ x)=\dfrac{y}{x}\\\\\implies \bf \dfrac{dy}{dx}(\dfrac{1-y.log\ x}{y})=\dfrac{y}{x}\\\\\implies \underbrace{\sf {\orange{\dfrac{dy}{dx}=\dfrac{y^2}{x(1-y.log\ x)}}}}$

$\bigstar$ More Info :

$\underline{\boxed{\bullet \;\;\bf \;log\ x^y=y.log\ x}}\\\\\underline{\boxed{\bullet \;\;\bf \dfrac{d}{dx}(log\ x)=\dfrac{1}{x}}}\\\\\underline{\boxed{\bullet \;\; \bf \dfrac{d}{dx}(uv)=v.\dfrac{du}{dx}+u.\dfrac{dv}{dx}}}$

Answer 2

CORRECT QUESTION :–

$\\ \sf {\huge{.}} \: if \: \: y = { {x}^{x} }^{x..... \infty } \: \: then \: \: prove \: \: that \: \: \dfrac{dy}{dx} = \dfrac{ {y}^{2} }{x(1 - y log(x)) }$

ANSWER :–

GIVEN :–

$\\ { \huge{.}}{ \bold{ \: \: \: y = { { {x}^{x} }^{x...... \infty } } }}$

TO PROVE :–

$\\ \to{ \bold{ \: \: \: \dfrac{dy}{dx} = \dfrac{ {y}^{2} }{x(1 - y log(x)) } }}$

SOLUTION :–

$\\ \implies{ \bold{ y = { { {x}^{x} }^{x...... \infty } } }}$

$\\ \implies{ \bold{ y = { { {x}^{(x} }^{x...... \infty )} } }}$

• We should write this as –

$\\ \implies{ \bold{ y = { { {x}^{(y)}} } }}$

• Take log on both side –

$\\ \implies{ \bold{ log(y) = log({x}^{y})}}$

• Using property –

$\\ \longrightarrow \: { \bold{ log({a}^{b}) =( b) log(a) }}$

• So –

$\\ \implies{ \bold{ log(y) = (y)log(x)}}$

• Now Differentiate with respect to 'x' –

$\\ \implies{ \bold{ \left( \dfrac{1}{y} \right) \dfrac{dy}{dx} = y\left( \dfrac{1}{x} \right) + log(x).\dfrac{dy}{dx}}}$

$\\ \implies{ \bold{ \left( \dfrac{1}{y} \right) \dfrac{dy}{dx} - log(x).\dfrac{dy}{dx} = y\left( \dfrac{1}{x} \right) }}$

$\\ \implies{ \bold{ \dfrac{dy}{dx} \left( \dfrac{1}{y} - log(x) \right) = \left( \dfrac{y}{x} \right) }}$

$\\ \implies{ \bold{ \dfrac{dy}{dx} \left( \dfrac{1 - y log(x) }{y} \right) = \left( \dfrac{y}{x} \right) }}$

$\\ \implies{ \bold{ \dfrac{dy}{dx} = \dfrac{ {y}^{2} }{x(1 - y log(x))}}}$

$\\ \: \: \: \: \: \: \: { \bold{ \underbrace{Hence \: \: proved}}}$

$\\ \rule{220}{2}$

USED FORMULA :–

$\\ { \bold{ (1) \: \: \dfrac{d( log(x)) }{dx} = \dfrac{1}{x} }}$

$\\ { \bold{ (2) \: \: \dfrac{d(u.v) }{dx} = u \dfrac{dv}{dx} + v \dfrac{du}{dx} }}$

$\\ \rule{220}{2}$