# Find the derivative of # x^2 + y^2 = (5x^2 + 4y^2 -x)^2 #, and find the equation of the tangent at #(0,0.25)#?

##### 1 Answer

The derivative is given (implicitly) by;

# x + ydy/dx = (5x^2 + 4y^2 -x)(10x+8ydy/dx-1) #

The equation of the tangent at

# y = x + 1/4 #

#### Explanation:

The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point.

When we differentiate

However, we cannot differentiate a non implicit function of

When this is done in situ it is known as implicit differentiation.

We have:

# x^2 + y^2 = (5x^2 + 4y^2 -x)^2 #

Differentiate wrt

# 2x+2ydy/dx = 2(5x^2 + 4y^2 -x)(10x+8ydy/dx-1) #

# :. x + ydy/dx = (5x^2 + 4y^2 -x)(10x+8ydy/dx-1) #

Whilst we could spend time and effort to find an explicit equation for

# => 0 + 1/4 dy/dx = (0 + 4(1/16) -0)(0+8/4dy/dx-1) #

# :. \ \ \ \ \ \ \ \ 1/4 dy/dx = (1/4)(2dy/dx-1) #

# :. \ \ \ \ \ \ \ \ 1/4 dy/dx = 1/2dy/dx-1/4 #

# :. \ \ \ \ \ \ \ \ 1/4 dy/dx = 1/4 #

# :. \ \ \ \ \ \ \ \ \ \ \ \ dy/dx = 1 #

So the tangent passes through

# \ \ \ \ \ y-1/4 = 1(x-0) #

# :. y-1/4 = x #

# :. \ \ \ \ \ \ \ \ y = x + 1/4 #

We can verify this solution graphically;

**Advanced Calculus**

There is another (often faster) approach using partial derivatives. Suppose we cannot find

# (partial F)/(partial x) (1) + (partial F)/(partial y) dy/dx = 0 => dy/dx = −((partial F)/(partial x)) / ((partial F)/(partial y)) #

So Let

#(partial F)/(partial x) = 2x - 2(5x^2 + 4y^2 -x)(10x-1)#

#(partial F)/(partial y) = 2y - 2(5x^2 + 4y^2 -x)(8y) #

And so:

# dy/dx = -(2x - 2(5x^2 + 4y^2 -x)(10x-1))/(2y - 2(5x^2 + 4y^2 -x)(8y))#

# \ \ \ \ \ \= -(x - (5x^2 + 4y^2 -x)(10x-1))/(y - 8y(5x^2 + 4y^2 -x))#

Here we get an immediate implicit function for the derivative, so again at

# dy/dx = -(0 - (0 + 4/16 -0)(0-1))/(1/4 - 8/4(0 + 4/16 -0))#

# \ \ \ \ \ \ = -(- (1/4)(-1))/(1/4 - 2(1/4))#

# \ \ \ \ \ \ = -(1/4)/(-1/4)#

# \ \ \ \ \ \ = 1 \ \ \ \ # , as before