Question

find dy/dx y =x^logx+logx^x​

Answer 1

EXPLANATION.

$\sf \implies \dfrac{dy}{dx} = x^{log(x)} + log(x)^{x}$

As we know that,

Let we assume that,

$\sf \implies x^{log(x)} = u$

$\sf \implies log(x)^{x} = v$

Now, we can write equation as,

$\sf \implies \dfrac{dy}{dx} = u + v$

Now first we differentiate the function.

$\sf \implies u = x^{log(x)}$

Taking log on both sides of the equation, we get.

$\sf \implies log(u) = log \bigg[ x ^{log(x)} \bigg]$

$\sf \implies log(u) = log(x)^{2}$

Differentiate both sides w.r.t x, we get.

$\sf \implies \dfrac{1}{u} . \dfrac{du}{dx} = 2 [log(x)]. \dfrac{d[log(x)]}{dx}$

$\sf \implies \dfrac{1}{u} . \dfrac{du}{dx} = 2 [log(x)] . \dfrac{1}{x}$

$\sf \implies \dfrac{du}{dx} = 2u \bigg[ \dfrac{log(x)}{x} \bigg]$

Put the value of u = x^(log x) in the equation, we get.

$\sf \implies \dfrac{du}{dx} = 2 [x^{log(x)} ] . \bigg[ \dfrac{log(x)}{x} \bigg]$

$\sf \implies \dfrac{du}{dx} = 2 \bigg(x^{log(x) - 1} \bigg) . log(x) . . . . . (1)$

Now we differentiate second function, we get.

$\sf \implies v = log(x)^{x}$

Taking log on both sides of the equation, we get.

$\sf \implies log(v) = log \bigg[ log(x)^{x} \bigg]$

Differentiate both sides w.r.t x, we get.

$\sf \implies \dfrac{1}{v} . \dfrac{dv}{dx} = x \bigg(log[log(x)]\bigg)$

$\sf \implies \dfrac{1}{v} . \dfrac{dv}{dx} = \dfrac{d(x)}{dx} \times [log(log(x))] \ + \ \dfrac{d[log(log(x))]}{dx} \times (x)$

$\sf \implies \dfrac{1}{v} . \dfrac{dv}{dx} = \bigg[ 1 \times log(log(x)) \ + (x) \times \dfrac{1}{log(x)} \times \dfrac{d(log(x))}{dx} \bigg]$

$\sf \implies \dfrac{1}{v} . \dfrac{dv}{dx} = \bigg[ log(log(x)) \ + \dfrac{x}{log(x)} \times \dfrac{1}{x} \bigg]$

$\sf \implies \dfrac{1}{v} . \dfrac{dv}{dx} = \bigg[ log(log(x)) + \dfrac{1}{log(x)} \bigg]$

$\sf \implies \dfrac{dv}{dx} = v \bigg[ log(log(x)) + \dfrac{1}{log(x)} \bigg]$

Put the value of v = log(x)ˣ in the equation, we get.

$\sf \implies \dfrac{dv}{dx} = log(x)^{x} \bigg[ log(log(x)) + \dfrac{1}{log(x)} \bigg] . . . . . (2)$

Now, we can write equation as,

$\sf \implies \dfrac{dy}{dx} = u + v$

$\sf \implies \dfrac{dy}{dx} = \dfrac{du}{dx} + \dfrac{dv}{dx}$

Put the values in the equation, we get.

$\sf \implies \dfrac{dy}{dx} = 2 \bigg( x^{log(x)- 1} \bigg) . log(x) \ + log(x)^{x} \bigg[ log(log(x)) + \dfrac{1}{log(x)} \bigg]$

Answer 2

Answer:

Given :-

• Find.

• $\frac{dy}{dx} = {x}^{log(x)} + log {x}^{x}$

To prove :-

• The value dy/dx.

Explanation :-

• Refer the given attachment for better understanding.
• It helps you better.

Hope it helps u mate .

Thank you .