# Calculating Derivative – Deriving an implicit function – Exercise 2122

Exercise

Given the equation:

$$\frac{y}{x^3}+\frac{x}{y^3}=x^2y^4$$

For

$$y=f(x)$$

Find the derivative of the following function:

$$y'=f'(x)$$

$$f'(x)=\frac{5x^4y^7-4x^3}{4y^3-7y^6x^5}$$

Solution

We need to calculate a derivative of an implicit function. To do that, we use the chain rule in Derivative Rules on both sides of the equation and when we encounter y, we derive it, i.e. multiply by y’. First, we get rid of the denominator:

$$\frac{y}{x^3}+\frac{x}{y^3}=x^2y^4 / \cdot x^3y^3$$

$$y^4+x^4=x^5y^7$$

Then we derive both sides by x:

$$4y^3y'+4x^3=x^5\cdot 7y^6y'+5x^4y^7$$

Notice that on the right side we used the multiplication rule in Derivative Rules.

Now, we want to isolate y’. Therefore, we move expressions with y’ to one side:

$$4y^3y'-7y^6x^5y'=5x^4y^7-4x^3$$

$$y'(4y^3-7y^6x^5)=5x^4y^7-4x^3$$

$$y'=\frac{5x^4y^7-4x^3}{4y^3-7y^6x^5}$$

Have a question? Found a mistake? – Write a comment below!
Was it helpful? You can buy me a cup of coffee here, which will make me very happy and will help me upload more solutions!

Share with Friends