- 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.

VIDEO LESSON


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 \(\displaystyle {dy\over dx}\).

VIDEO LESSON


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.
\(\displaystyle 6x+2y {dy\over dx}=4\)
\(\displaystyle 2y {dy\over dx}=4-6x\)
\(\displaystyle {dy\over dx}={4-6x \over 2y}\)
\(\displaystyle {dy\over dx}={2-3x \over y}\)

VIDEO LESSON


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.
\(\displaystyle 1+{dy\over dx}={10x\over 4(x+y)^3}\)
\(\displaystyle {dy\over dx}={10x\over 4(x+y)^3-1}\)

VIDEO LESSON


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.
\(\displaystyle x^2 {dy\over dx}=3-2xy\)
\(\displaystyle {dy\over dx}={3-2xy\over x^2}\)

VIDEO LESSON

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.
Gradient at (6, -1)
\(\displaystyle {dy\over dx}={6-2(6)\over 2(-1)-2}={3\over2}\)
  • 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.
Gradient at (6, 3)
\(\displaystyle {dy\over dx}={6-2(6)\over 2(3)-2}=-{3\over2}\)
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

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