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## Implicit Differentiation

Q:  Using implicit differentiation, find dy/dx of

y/(x+6y)=x7-9

at the point (1,-8/49)

A:

Taking the derivative of the right side of the equation is fairly basic, taking the derivative of the left side, on the other hand, is harder.  The left side involves a quotient rule.  I am going to ignore the right side of the equation for now and just deal with the left:

y/(x+6y)

The quotient rule tells you (in some form / notation or another):

(bottom)*(deriv. of top) – (deriv of bottom)*(top) all divided by (bottom) squared.  [You may have learned this with f’s and g’s and f ‘ and g ‘ — it all says the same thing]

So:

Top = y

Bottom = x + 6y

Derivative of Top = dy/dx = y’ [whatever notation you are using]

Derivative of Bottom = 1 + 6(dy/dx) = 1 + 6y’

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## Related Rates

Q:  If y3=16x2, find dx/dt when x=4 and dy/dt=-1

A:

First thing is to take the derivative of the equation implicitly to get:

(3y2)dy/dt = (32x)dx/dt

The goal is to solve for dx/dt… So we need to know y, x, and dy/dt.  We know x and dy/dt (given in the problem).  Use the original equation to solve for y when x is 4:

y3=16(4)2

which gives:

y = cuberoot(256)

So, now we know all that we need to know to solve for dx/dt:

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## Implicit Differentiation

Q:

A.  Implicitly find dy/dx of exy=8

B.  Now, solve exy=8 for y first, then take the derivative.  Compare your answers to A and B.

A:

A.  Remember, implicit differentiation is just the chain rule:  y is a function of x.  So, we need to use the product rule since we are multiplying two functions ex times y:

Also remember, the derivative of x with respect to x is 1 and the derivative of y with respect to x is dy/dx:

Differentiate as follows:

exy + ex(dy/dx) = 0

Now, solve for dy/dx:

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## Implicit Differentiation Example

Q:  Find y'(3) using implicit differentiation of the equation:

2x2 + 3x + xy = 3

and y(3) = -8

A:

First, take the derivative of the equation (this will involve a product rule on the “xy” term):

The implicit differentiation of 2x2 + 3x + xy = 3 is (the product rule is used to differentiate the term “xy”):

4x + 3 + xy’ + y = 0

And we know that when x = 3, y = -8.  So, plug that in and the solve for y’:

4(3) + 3 + 3y’ + (-8) = 0

12 + 3 + 3y’ – 8 = 0

7 + 3y’ = 0

3y’ = -7

y’ = -7/3

So, y'(3) = -7/3