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