Question

find du/dx when u=xlogxy where x^3+y^3+3xy=1

Answer 1

Step-by-step explanation:

du/dx

u= xlogy

on putting the value of u

hope u undetstand....

Answer 2

$\dfrac{du}{dx} =\dfrac{x}{y}(-\frac{x^{2}}{y^{2}+x})+\dfrac{1}{x} +\log xy$

Step-by-step explanation:

We have,

$u=\loog xy$

Differentiating w.r,t, x, we get

$\dfrac{du}{dx} =x\dfrac{1}{xy} (x\dfrac{dy}{dx}+y)+\log xy$

$\dfrac{du}{dx} =\dfrac{x}{y}\dfrac{dy}{dx}+\dfrac{1}{x} +\log xy$   .....(1)

Also,

$x^3+y^3+3xy=1$

∴$3x^2+3y^2\dfrac{dy}{dx} +3x\dfrac{dy}{dx} =0$

⇒ $\dfrac{dy}{dx}=-\frac{3x^{2}}{3y^{2}+3x} =-\frac{x^{2}}{y^{2}+x}$

Putting the value of $\dfrac{dy}{dx}$ in (1), we get

$\dfrac{du}{dx} =\dfrac{x}{y}(-\frac{x^{2}}{y^{2}+x})+\dfrac{1}{x} +\log xy$

Hence, $\dfrac{du}{dx} =\dfrac{x}{y}(-\frac{x^{2}}{y^{2}+x})+\dfrac{1}{x} +\log xy$.