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Implicit Differentiation

Subject:

Calculus [1]

Subject X2:

Calculus [1]

Implicit Differentiation

So far, all the functions being differentiated are explicit functions, meaning that one of the variables was specifically given in terms of the other variable.

ex.

*f*(x) = 3x^{2} + 1 then f '(x) = 6x

However, not all functions are given explicitly and are only implied by an equation.

ex.

xy = 1 is an equation given implicitly, explicitly it is y = 1/ x. Now to find *dy/dx* for xy = 1, simply solve for y and differentiate.

xy = 1

y = 1 / x = x^{-1}*dy/dx* = -1 x^{-2}= 1/x^{2}

But, not all equations are easily solved for y, as in the equation 3x + y^{3 }= y^{2 }+ 4, This is where implicit differentiation is applied. Implicit differentiation is taking the derivative of both sides of the equation with respect to one of the variables. Most commonly, used is the derivative of y with respect to x. or *dy/dx*. Since we have not solved for y as a function of x, the derivative of y must be left as* dy/dx.*

From the example above: 3x + y^{3 }= y^{2 }+ 4, solve for* dy/dx*.

3x + y^{3 }= y^{2 }+ 4*d/dx*(3x + y^{3}) = *d/dx*(y^{2} + 4)

3 + 3y^{2 }*dy/dx* = 2y *dy/dx*

3 = 2y *dy/dx* - 3y^{2 }*dy/dx*

3 = y ( 2 - 3y ) *dy/dx*

3 / y (2 - 3y ) = *dy/dx*

Finding the slope of a curve implicitly:

ex.

find the slope of the curve x^{2 }+ y^{3 }= 2x + y at ( 2,4)

*d/dx* [x^{2} + y^{3}] = *d/dx *[2x + y]

2x + 3y^{2 }*dy/dx* = 2 + *dy/dx*

2x - 2 = (-3y^{2} + 1) *dy/dx*

2( x - 1) / (-3y^{2} + 1) = *dy/dx *= slope of the curve

substitute (2,4) into dy/dx to find the slope at that point.

2(2-1) / (-3 · 4^{2} + 1) = 2 /-49 = -2/49 is the slope of the curve.

Subject:

Calculus [1]

Subject X2:

Calculus [1]

**Links:**

[1] http://www.course-notes.org/Subject/Math/Calculus