Question

What is the derivative of \[XY\] ?

Answer 1

Hint:To find the derivative we can use the slope formula, that is \[slope=\dfrac{dy}{dx}\]. \[dy\] is the changes in \[Y\]and \[dx\] is the changes in \[X\].
 

To find the derivative of two variables in multiplication, this formula is used
\[\dfrac{d}{dx}(uv)=u\dfrac{d}{dx}(v)+v\dfrac{d}{dx}(u)\]
Mechanically, \[\dfrac{d}{dx}(f(x))\] measures the rate of change of \[f(x)\] with respect to \[x\].Differentiation of any constant is zero. Differentiation of constant and a function is equal to constant times the differentiation of the function.

Complete step by step answer:
Let us derivate \[XY\] with respect to \[X\]. Using the derivative formula,
\[\dfrac{d}{dx}(uv)=u\dfrac{d}{dx}(v)+v\dfrac{d}{dx}(u)\]
Substituting the terms
\[\dfrac{d}{dX}(XY)=X\dfrac{d}{dX}Y+Y\dfrac{d}{dX}X\]
The differentiation of \[X\] with respect to \[X\] is given by
\[\dfrac{d}{dX}X=1\]
As the derivation of the function is with respect to \[X\] , the variable \[Y\] cannot be differentiable. \[Y\] is not constant.
The derivative of \[XY\] with respect to \[X\] is
\[\dfrac{d}{dX}(XY)=X\dfrac{d}{dX}Y+Y\]

Let us derivate \[XY\] with respect to \[Y\].Using the derivative formula,
\[\dfrac{d}{dx}(uv)=u\dfrac{d}{dx}(v)+v\dfrac{d}{dx}(u)\]
Substituting the terms
\[\dfrac{d}{dY}(XY)=X\dfrac{d}{dY}Y+Y\dfrac{d}{dY}X\]
The differentiation of \[Y\] with respect to \[Y\] is given by
\[\dfrac{d}{dY}Y=1\]
As the derivation of the function is with respect to \[Y\], the variable \[X\] cannot be differentiable. \[X\] is not constant.
The derivative of \[XY\] with respect to \[Y\] is
\[\therefore \dfrac{d}{dY}(XY)=X+Y\dfrac{d}{dY}X\]

Note: \[\dfrac{d}{dx}(f(x))=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(x+h)-f(x)}{h}\] is the formula for finding the derivative from the first principles. The slope is the rate of change of \[y\] with respect to \[x\] that means if \[x\] is increased by an additional unit the change in \[y\] is given by \[\dfrac{dy}{dx}\] . Let us understand with an example, the rate of change of displacement of an object is defined as the velocity \[Km/hr~\] that means when time is increased by one hour the displacement changes by \[Km\]. For solving derivative problems different techniques of differentiation must be known.