Solveeit Logo
Login

Question

Question: How do you implicitly differentiate \[{x^2}{y^3} - xy = 10\]?...

How do you implicitly differentiate x2y3−xy=10{x^2}{y^3} - xy = 10?

Explanation

Solution

According to the question, we will start with normal differentiation. We will differentiate it with respect to ‘x’. After that, we will use the product rule and the chain rule. At last, we will try to rearrange the terms, simplify them, and solve them to get the answer.

Complete step-by-step solution:
The given equation is:
x2y3−xy=10{x^2}{y^3} - xy = 10
First, we will try to differentiate both left and right sides, with respect to xx, and we will get:
D(x2y3−xy)=D(10)D({x^2}{y^3} - xy) = D(10)
Now, we will try to apply the product rule and chain rule here, and we get:
⇒x23y2y′+y32x−(xy′+y)=0\Rightarrow {x^2}3{y^2}y' + {y^3}2x - (xy' + y) = 0
Now, we will pen the brackets, and we get:
⇒x23y2y′+y32x−xy′−y=0\Rightarrow {x^2}3{y^2}y' + {y^3}2x - xy' - y = 0
Now, we will rearrange the terms, and we get:
⇒3x2y2y′+2y3x−xy′−y=0\Rightarrow 3{x^2}{y^2}y' + 2{y^3}x - xy' - y = 0
Now, we will try to shift some terms to the other side. We will shift the y′y'terms on the left side, and shift all other terms to the right side, and we get:
⇒3x2y2y′−xy′=−2y3x+y\Rightarrow 3{x^2}{y^2}y' - xy' = - 2{y^3}x + y
Now, we will try to take out the common terms from both the sides. On left side, the common term is y′y', and on right side, the common term is yy:
⇒y′(3x2y2−x)=y(1−2xy2)\Rightarrow y'(3{x^2}{y^2} - x) = y(1 - 2x{y^2})
Now, we will try to keep y′y'alone. We will shift all the terms related to y′y'on the other side of the equation. Here the term 3x2y2−x3{x^2}{y^2} - xgets divided when it goes to the other side of the equation, and we get:
⇒y′=dfracy(1−2xy2)(3x2y2−x)\Rightarrow y' = dfrac{{y(1 - 2x{y^2})}}{{(3{x^2}{y^2} - x)}}
Now, we will take out the common term xxfrom the denominator, and we get:
⇒y′=dfracy(1−2xy2)x(3xy2−1)\Rightarrow y' = dfrac{{y(1 - 2x{y^2})}}{{x(3x{y^2} - 1)}}

Therefore, the final answer is y′=dfracy(1−2xy2)x(3xy2−1) y' = dfrac{{y(1 - 2x{y^2})}}{{x(3x{y^2} - 1)}}

Note: Implicit differentiation is a type of method where differentiation takes place when we are having a function which is in terms of xxas well as of yy. If we take an example, x2+y2=16{x^2} + {y^2} = 16, then we can tell that this is the formula of a circle, which is having a radius 44. After that we do the implicit differentiation.