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Implicit Differentiation xy

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When we talk about Implicit Differentiation xy, it means we are differentiating both the Functions x and y. When we learn normal differentiation, we use to do differentiation with respect to one variable and all the other variables are constant. For example, if we are doing differentiation with respect to x then, all the other variables are treated as constant. But, in implicit differentiation xy we differentiate both the functions x as well as y, for doing implicit differentiation xy we need to follow some basic steps, they are given as:

Step 1: Differentiate both sides, with respect to 'x'.

Step 2: Send all the terms containing y one side of the equation, and all terms without to the other side.

Step 3: On the side with 'y' terms, do the differentiation.

When we solve any problem related to implicit differentiation xy the above mentioned process is followed. Now, have a look on few examples in order to understand the topic.

Example 1: Assume y is a function of x, find dy/dx of 3x-y = 4?

For solving this type of problem we need to take the derivative of both the side and of all the variables.

=> d/dx(3x)- d/dx(y)= d/dx(4),

=> 3-dy/dx=0,

=> dy/dx= 3,

This is the required solution for given equation.

Note: Whenever we talk about implicit differentiation xy then, it will take only xy as variable all the other terms are constant.

Example 2: Assume y is a function of x .find dy/dx for (x-y) 2 =x+y-12?

Solution: We will take the derivative of both the side then, we will apply Chain Rule.
=> d/dx((x-y) 2) = d/dx(x+y-12),
=> d/dx(X2 +y2 -2xy) = d/dx(x+y+12,)
For integrating 2xy we need to apply Product rule, as we have differentiating terms in multiplication.
=> 2x +2y*dy/dx -2(xdy/dx +y(1)) = 1 +dy/dx,
=> 2x +2ydy/dx -2xdy/dx-2y = 1+dy/dx,
=> dy/dx(2y-2x-1) = 1-2x-2y,
=> dy /dx = 1-2x-2y /2y-2x-1.
This is the required solution.

Example 3: Assume y is the function of x differentiate y= cos(3x+3y)?

Solution: For solving this problem you must have the knowledge of differentiation of Trigonometric Functions then only you will be able to solve this type of problem. Now, talk about the solution of the problem.

Firstly, we take the derivative of both the side then, we will differentiate the function,

=> d/dx(y) = d/dx(cos(3x+3y)),

Now, we need to apply chain rule for solving this problem,
=> dy/dx = sin(3x +3y) *(3 +3dydx),
=> dy/dx =-3sin(3x+3y) -3sin(3x+3y)dy/dx,
=> dy/dx(1-3sn(3x+3y)) = 3sin(3x+3y),
=> dy/dx = 3sin(3x+3y) /1-3sin(3x+3y).
This is the required solution for the given problem.

Example 4: Assume y is a function of x, differentiate the following function:

Y2 +4x +y = ln x?

Solution:

As you are seeing that a new function is introduced and we know that ln x is differentiated as 1/x now, it’s simple to solve this problem we need to follow the basic steps only.

Firstly, we will take the derivative of both the side then, we will apply chain rule:
=> d/dx (Y2 +4x +y) = d/dx(ln x),
=> 2y *dy/dx + 4 +6dy/dx =1/x,
=> dy/dx(2y +1) = (1/x) -4,
=> dy/dx =((1/x) -4) /(2y +1),

This is the required answer.

Now, we will move to some examples where we need to find the value of the function at particular points, in this type of questions, the value of x and y is known and we have to find the value of derivative at that particular Point. Now, we will see some example then, the whole concept will be clear.

Example 5: Find dy/dx at point (1,1) if x2 +4y2 =0?

Solution: We need to take the derivative of both the side first then, we will differentiate.

=> d/dx( x2) + d/dx(4y2) =0,

=>2x + 8y* dy/dx = 0,

=> 8y* dy/dx = -2x,

=> dy/dx = -2x/8y,

The points (1,1) tells us that value of x is 1 and value of y is 1. so,

=> dy/dx = -2*1/8*1,

=> dy/dx = -1/4,

This is the required answer.

Example 6: Find dy/dx at point (-1,1) if x2 +4y2 +x3 +y=0?

Solution: We need to take the derivative of both the side first then, we will integrate them.

=> d/dx(x2) +d/dx(4y2) +d/dx(x3) + d/dx(y) =0,

=> 2x +8y (dy/dx) +3x2 +dy/dx =0,

Now, we will separate the terms as:

=> 8y(dy/dx) +dy/dx =-2x -3x2,

=> dy/dx(1+8y) = -2x-3x2,

=> dy/dx= -2x-3x2/1+8y,

Now, put the given point to get the required solution,

We will put x=-1 and y=1,

=> -2(-1) -3((-1)2) / 8*1,

=> -1/8.

This is the required solution.

The point (-1,1 ) shows the value at the Slope of given Tangent and -1/8 is the point where the given function satisfied both the given point.

Example 7: Find dy/dx at point(0,0) if, x2 +4y2 +cosx +y=0?

Solution: We need to take the derivative of both the side first then, we will integrate them.

=> d/dx(x2) +d/dx(4y2) +d/dx(cosx) + d/dx(y) =0,

=> 2x +8y (dy/dx) -sinx +dy/dx =0,

Now, we will separate the terms,

=> 8y(dy/dx) +dy/dx =-2x +sinx,

=> dy/dx(1+8y) = -2x+sinx,

=> dy/dx = -2x+ sinx /1+8y,

Now, putting the given points in the equation we will get:

=> dy/dx = -2(0) + sin0 /1+8(0),

As we know, sin0 =0 so, the value of the function at (0,0) will be 0.

This is the required answer.

Example 8: Find dy/dx at points (90,0) if ysinx +cosx =0?

Solution: We need to take the derivative of both the side first then, we will differentiate them.

=> d/dx(ysinx) + d/dx(cosx),

Now, we need to apply product rule,

=> y *d/dx(sinx) + d/dx(y) siinx =0,

=> y* cosx + dy/dx * sinx =0,

=> dy/dx = -y*cosx /sinx –sinx,

Now, put the given point to get the required solution,

Hence, y=0 so, whole equation will be 0.

So, we will remain with –sin90 =-1.

Example 9: Find dy/dx at points (90,0) if ysinx +xcosx =0?

Solution: We need to take the derivative of both the side first then, we will differentiate them.

=> d/dx(ysinx) + d/dx(xcosx),

Now, we need to apply product rule for both the functions,

=> y d/dx(sinx) + d/dx(y) siinx + xd/dx(cosx) +cosx*1 =0,

=> y* cosx + dy/dx * sinx - xsinx +cosx =0,

=> dy/dx = xsinx – cosx - ycosx/sinx,

As we know sin90=1 and cos 90=0. On putting the values we will get:

=> dy/dx = x*1-0-0/1,

=> dy/dx = 90,

This is the required solution for the question. In this way, you can easily solve the problems related to implicit differentiation xy.