6.2.1 Using Implicit Differentiation Flashcards
Using Implicit Differentiation
• Find the derivative of a relation by differentiating each side of its equation implicitly and solving for the derivative as an unknown. This process is called implicit differentiation.
note
- Implicit differentiation often produces a derivative expressed in terms of more than one variable. When evaluating the slope of a line tangent to a point of a relation, it is necessary to substitute both the x-value and the y-value of the point into the derivative.
- Notice that you could substitute any values for x and y into the derivative. However, only ordered pairs of the original relation produce reasonable answers.
- Here is a complicated-looking relation.
- Find the derivative implicitly by taking the derivative of both sides of the implicit equation.
- Now you can differentiate each term piece by piece.
- Sometimes you will have to use different differentiation rules in the middle of a problem. Here the product and chain rules are both used.
- Once you have differentiated each term, you can solve
for dy/dx.
Given the equation 1−ln(xy)=e^y,find dydx.
dy/dx=−y/xye^y+x
Given the equation xy=5,find dy/dx.
dy/dx=−y/x
Given the equation sinxy=1/2,find dy/dx.
dy/dx=−y/x
Given the equation x^2y+y^2x=0,find dy/dx.
dy/dx=−2xy−y^2/2xy+x^2
Given x/y=1/9,find dy/dx.
dy/dx=9
Suppose a curve is defined by the equation (6−x)y^2=x^3. What is the equation of the line tangent to the curve at (3, 3)?
y=2x−3
Given the equation2x+2y+xy^2=5,find dy/dx.
dy/dx=−2+y^−2/2−2xy^−3
Suppose a curve is defined by the equation y^2=x^3(2−x). What is the equation of the line tangent to the curve at (1,−1)?
y=−x
Given the equation x^2+3x=y^2+y−6,find dy/dx.
dy/dx=2x+3/2y+1
Given cos^3(sinxy)=k where k is some constant,find dxdy
dx/dy=−x/y
Given the equation xy=cotxy,find dy/dx.
dy/dx=−y/x
Given sin^3(cosxy)=k where k is some constant, find dy/dx.
dy/dx=−y/x
