## - Implicit Differentiation

A function which is written in the form $$y=f(x)$$ is called an explicit function: $$y$$ is stated explicitly in terms of $$x$$. However, functions such as $$x^2+y^2=0$$ or $$\displaystyle xy+{1\over x}=5x^2$$ are implicit functions.

We will now look at how to differentiate a function in one variable with respect to another variable.

LESSON 1
Differentiate $$y^2$$ with respect to $$x$$.
SOLUTION
$$\displaystyle {d\over dx} y^2$$
• We treat $$\displaystyle {d\over dx}$$ as a fraction and separate the numerator and the denominator.
$$\displaystyle ⟹{d\over \:} y^2 {\: \over dx}$$
• We multiply the expression by $$\displaystyle {dy\over dy}$$ since this term is equivalent to 1.
$$\displaystyle ⟹{d\over dy} y^2 {dy\over dx}$$
$$\displaystyle ⟹2y {dy\over dx}$$
To differentiate a function in $$y$$ with respect to $$x$$ we differentiate with respect to $$y$$ and then multiply by $$\displaystyle {dy\over dx}$$.

LESSON 2
Use implicit differentiation to determine $$\displaystyle {dy\over dx}$$ for $$3x^2+y^2=4x$$.
SOLUTION
$$3x^2+y^2=4x$$
Differentiate each term with respect to $$x$$.
$$\displaystyle {d\over dx} (3x)^2+{d\over dx}y^2={d\over dx} (4x)$$
We will now rearrange to make $$\displaystyle {dy\over dx}$$ the subject.
$$\displaystyle 6x+2y {dy\over dx}=4$$
$$\displaystyle 2y {dy\over dx}=4-6x$$
$$\displaystyle {dy\over dx}={4-6x \over 2y}$$
$$\displaystyle {dy\over dx}={2-3x \over y}$$

LESSON 3
Determine $$\displaystyle {dy\over dx}$$ for $$(x+y)^4-5x^2=0$$.
SOLUTION
$$(x+y)^4-5x^2=0$$
• Differentiate using the CHAIN RULE.
$$\displaystyle (x+y)^3 (1+{dy\over dx})-10x=0$$
• Make $$\displaystyle {dy\over dx}$$ the subject.
$$\displaystyle 1+{dy\over dx}={10x\over 4(x+y)^3}$$
$$\displaystyle {dy\over dx}={10x\over 4(x+y)^3-1}$$

LESSON 4
Determine $$\displaystyle {dy\over dx}$$ for $$x^2 y-3x=5$$.
SOLUTION
$$\displaystyle x^2 y-3x=5$$
• Differentiate using the PRODUCT RULE.
$$\displaystyle 2xy+x^2 {dy\over dx}-3=0$$
• Make $$\displaystyle {dy\over dx}$$ the subject.
$$\displaystyle x^2 {dy\over dx}=3-2xy$$
$$\displaystyle {dy\over dx}={3-2xy\over x^2}$$

# APPLICATIONS OF IMPLICIT DIFFERENTIATION

LESSON 1
Find the equations of the tangents at the points where $$x=6$$ on the curve $$x^2+y^2-6x-2y=3$$.
SOLUTION
• Differentiate implicitly to find an expression for $$\displaystyle {dy\over dx}$$.
$$x^2+y^2-6x-2y=3$$
$$\displaystyle 2x+2y {dy\over dx}-6-2 {dy\over dx}=0$$
$$\displaystyle 2y {dy\over dx}-2 {dy\over dx}=6-2x$$
$$\displaystyle (2y-2) {dy\over dx}=6-2x$$
$$\displaystyle {dy\over dx}={6-2x \over 2y-2}$$
• Substitute the value of $$x$$ into the original equation to determine the corresponding value(s) for $$y$$.
When $$x=6$$
$$6^2+y^2-6(6)-2y=3$$
$$y^2-2y-3=0$$
$$(y-3)(y+1)=0$$
$$y=-1, 3$$
• Determine the gradient of the tangent.
$$\displaystyle {dy\over dx}={6-2(6)\over 2(-1)-2}={3\over2}$$
• Determine the equation of the tangent.
Equation of line: $$y=mx+c$$
Using (6, -1)
$$\displaystyle -1={3\over2} (6)+c$$
$$-10=c$$
$$\displaystyle y={3\over2} x-10$$
• We have the equation of one tangent so we need to determine the equation of the second tangent.
$$\displaystyle {dy\over dx}={6-2(6)\over 2(3)-2}=-{3\over2}$$
Equation of line
Using (6, 3)
$$\displaystyle 3=-{3\over2} (6)+c$$
$$12=c$$
$$\displaystyle y=-{3\over2} x+12$$

LESSON 2
Find and classify the stationary points on the curve $$2xy+y-x^2=6$$.
SOLUTION
• We need to determine $$\displaystyle {dy\over dx}$$ by differentiating implicitly.
$$2xy+y-x^2=6$$
$$2y+2x {dy\over dx}+{dy\over dx}-2x=0$$
$$(2x+1) {dy\over dx}=2x-2y$$
$$\displaystyle {dy\over dx}={2x-2y\over 2x+1}$$
• Stationary points occur when $$\displaystyle {dy\over dx}=0$$.
$$\displaystyle {2x-2y \over 2x+1}=0$$
$$2x-2y=0$$
$$x=y$$
• We now know that at the stationary points $$x=y$$. Therefore, we can sub. $$x=y$$ into original equation.
$$2x(x)+x-x^2-6=0$$
$$x^2+x-6=0$$
$$(x+3)(x-2)=0$$
$$x=-3, 2$$
$$y=-3,2$$
(-3,-3) and (2,2)
To determine the nature of the stationary points the second derivative needs to be evaluated at the values of $$x$$ which correspond to the stationary points.
• We differentiate $$\displaystyle {dy\over dx}={2x-2y \over 2x+1}$$ implicitly using the QUOTIENT RULE.
$$\displaystyle {d^2y \over dx^2}={(2-2 {dy\over dx})(2x+1)-(2x-2y)(2)\over (2x+1)^2}$$
NB: We have already stated that $$\displaystyle {dy\over dx}=0$$ and $$2x-2y=0$$.
$$\displaystyle {d^2y\over dx^2}={2(2x+1)\over (2x+1)^2}={2\over (2x+1)}$$
For $$(-3, -3)$$
$$\displaystyle {d^2y\over dx^2}={2\over 2(-3)+1}$$
$$\displaystyle {d^2y\over dx^2}=-{2\over5}$$ →Maximum

For (2, 2)
$$\displaystyle {d^2 y\over dx^2}={2\over 2(2)+1}$$
$$\displaystyle {d^2y\over dx^2}={2\over5}$$ →Minimum

Discussion