Question

Find dy/dx if y= log2^x + logx^x
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Answer 1

Step-by-step explanation:

मौसम सर्द हो या लहज़ा मेरी दोनों से ही नहीं बनती

Answer 2

Given :

$\bf y = log {2}^{x} + log \: {x}^{x}$

To find :

$\bf \dfrac{dy}{dx}$

Solution :

$\sf \implies y = log {2}^{x} + log \: {x}^{x}$

We know that :-

$\underline{ \boxed{ \bf log \: {a}^{n} = n \: log \: a }}$

$\sf \implies y =x \: log {2}+x \: log \: {x}$

$\sf \implies \dfrac{dy}{dx} = \dfrac{d(x \: log {2}+x \: log \: {x} )}{dx}$

$\sf \implies \dfrac{dy}{dx} = \dfrac{d(x \: log {2} )}{dx} + \dfrac{d(x \: log \: {x} )}{dx}$

$\sf \implies \dfrac{dy}{dx} =log {2} \dfrac{d(x \: )}{dx} + \dfrac{d(x \: log \: {x} )}{dx}$

$\sf \implies \dfrac{dy}{dx} =log {2} + \dfrac{d(x \: log \: {x} )}{dx}$

Now to differentiate xlogx use product rule :-

$\boxed{ \bf\dfrac{d(uv)}{dx} = u \dfrac{dv}{dx} + v u \dfrac{du}{dx}}$

$\sf \implies \dfrac{dy}{dx} =log {2} + \dfrac{d(x \: )}{dx}logx +x \dfrac{d(log \: x \: )}{dx}$

We know that :-

$\boxed{ \bf\dfrac{d(log \: t)}{dt} = \dfrac{1}{t} }$

$\sf \implies \dfrac{dy}{dx} =log {2} + logx + \bigg(x \times \dfrac{ \: 1}{x}\bigg)$

$\sf \implies \dfrac{dy}{dx} =log {2} + logx + 1$

We know that :-

$\boxed{ \bf \large log \: {a} + log \: b = log \: ab}$

$\sf \implies \dfrac{dy}{dx} =log {2x} + 1$

Answer :

$\large \bf \: log {2x} + 1$