Implicit Differentiation

Key Questions

  • How do you find the second derivative by implicit differentiation?

    Let us find {d^2y}/{dx^2} for x^3+y^3=1.

    First, let us find {dy}/{dx}.
    x^3+y^3=1
    by differentiating with respect to x,
    Rightarrow 3x^2+3y^2{dy}/{dx}=0
    by subtracting 3x^2,
    Rightarrow3y^2{dy}/{dx}=-3x^2
    by dividing by 3y^2,
    Rightarrow {dy}/{dx}=-{x^2}/{y^2}

    Now, let us find {d^2y}/{dx^2}.
    by differentiating with respect to x,
    Rightarrow{d^2y}/{dx^2}=-{2x cdot y^2-x^2 cdot 2y{dy}/{dx}}/{(y^2)^2} =-{2x(y^2-xy{dy}/{dx})}/{y^4}
    by plugging in {dy}/{dx}=-{x^2}/{y^2},
    Rightarrow{d^2y}/{dx^2}=-{2x[y^2-xy(-x^2/y^2)]}/y^4=-{2x(y^2+x^3/y)}/y^4
    by multiplying the numerator and the denominator by y,
    Rightarrow{d^2y}/{dx^2}=-{2x(y^3+x^3)}/y^5
    by plugging in y^3+x^3=1,
    Rightarrow{d^2y}/{dx^2}=-{2x}/y^5

    Wataru · 1 · Sep 12 2014
  • How do you find the second derivative by implicit differentiation on x^3y^3=8 ?

    As the first step, we will differentiate once, and apply the product rule:

    d/dx[x^3]*y^3 + d/dx[y^3]*x^3 = d/dx[8]

    For y^3, remember to use the chain rule. Simplifying yields:

    3x^2y^3 + 3y^2x^3dy/dx = 0

    Now, we will solve for dy/dx:

    dy/dx = -(3x^2y^3)/(3y^2x^3)

    We can cancel off the 3, an x^2, and a y^2, which will yield:

    dy/dx = -y/x

    Now, differentiate once again. We will apply the quotient rule:

    (d^2y)/(dx^2) = -(x*dy/dx - y*1)/x^2

    Looking back at the previous equation for dy/dx, we can substitute into our equation for the second derivative to get it in terms of only x and y:

    (d^2y)/(dx^2) = -(x*(-y/x) - y*1)/x^2

    Simplifying yields:

    (d^2y)/(dx^2) = (2y)/x^2

    Jacob F. · 1 · Jul 31 2014
  • What is implicit differentiation?

    Implicit differentiation is a way of differentiating when you have a function in terms of both x and y. For example:

    x^2+y^2=16

    This is the formula for a circle with a centre at (0,0) and a radius of 4

    So using normal differentiation rules x^2 and 16 are differentiable if we are differentiating with respect to x

    d/dx(x^2)+d/dx(y^2)=d/dx(16)

    2x+d/dx(y^2)=0

    To find d/dx(y^2) we use the chain rule:

    d/dx=d/dy *dy/dx

    d/dy(y^2)=2y*dy/dx

    2x+2y*dy/dx=0

    Rearrange for dy/dx

    dy/dx=(-2x)/(2y

    dy/dx=-x/y

    So essentially to use implicit differentiation you treat y the same as an x and when you differentiate it you multiply be dy/dx

    Youtube Implicit Differentiation

    Plmz1221 · 2 · Aug 4 2014

Questions

Basic Differentiation Rules