Question

differentiate u=xy +yz +zx with respect to x if y and z are constant terms​

Answer 1

Given

u = xy+yz+zx

where, z and y are constant terms

To Find

Derivative w.r.t. x

Solution

The equation is,

u = xy+yz+zx

Differentiating on both sides w.r.t. x

$\mathsf{\dfrac{du}{dx}= \dfrac{d[xy+yz+zx]}{dx}}$

$\mathsf{\dfrac{du}{dx}= \dfrac{d(xy)}{dx} +\dfrac{d(yz)}{dx} + \dfrac{d(zx)}{dx}}$

$\mathsf{\dfrac{du}{dx}= y \times\dfrac{dx}{dx} + 0 + z \times \dfrac{dx}{dx}}$

$\mathsf{\dfrac{du}{dx}= y +z }$

Answer 2

Answer:

Concept: using the formulae of derivatives for the answers.

Given: an equation $u=xy+yz+zx$

To find: differentiate the equation

Step-by-step explanation:

given an equation ,

let's differentiate the equation on both sides w.r.t to x.

now, $\frac{d}{dx}= \frac{d(xy+yz+zx)}{dx}$

so now,  separating  $\frac{d}{dx}$   based on addition , we will get

$\frac{d}{dx}= \frac{d(xy)}{dx}+\frac{d(yz)}{dx}+\frac{d(zx)}{dx}$

so, $\frac{d}{dx}= y\frac{d(x)}{dx}+0+z\frac{d(x)}{dx}$

so, $\frac{d}{dx}= y+z$.

∴ the answer is $\frac{d}{dx}= y+z$.

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