# Implicit Differentiation

• May 25th 2010, 08:47 AM
krzyrice
Implicit Differentiation
Assuming that the equations define x and y implicitly as differentiable functions x = f(t), y = g(t), find the slope of the curve x = f(t),
y = g(t) at the given value of t.

http://www.testdesigner.com/include/...-645138696.png

i tried to differential both functions but there are still y and x-variables. can someone guide me through this?
• May 25th 2010, 09:37 AM
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Quote:

Originally Posted by krzyrice
Assuming that the equations define x and y implicitly as differentiable functions x = f(t), y = g(t), find the slope of the curve x = f(t),
y = g(t) at the given value of t.

http://www.testdesigner.com/include/...-645138696.png

i tried to differential both functions but there are still y and x-variables. can someone guide me through this?

Using the first equation, plug in t = 0 and you will get x = 0.

Using the second equation, plug in t = 0 and you will get y = 9.

So when you differentiate implicitly for the first equation, you will have some x's, t's, and dx/dt's. Just solve for dx/dt and don't plug in values yet.

Then for the second equation you will have y's, t's, and dy/dt's. Solve for dy/dt without plugging in values.

Then take dy/dt divided by dx/dt to get dy/dx.

Then plug in x = 0, y = 9, and t = 0 and you should get the proper answer.
• May 25th 2010, 09:38 AM
AllanCuz
Quote:

Originally Posted by krzyrice
Assuming that the equations define x and y implicitly as differentiable functions x = f(t), y = g(t), find the slope of the curve x = f(t),
y = g(t) at the given value of t.

http://www.testdesigner.com/include/...-645138696.png

i tried to differential both functions but there are still y and x-variables. can someone guide me through this?

This is okay, consider the following

$\displaystyle y=f(x)= e^{x+5}$

And

$\displaystyle lny = lne^{x+5} = x+5$

If we want to differentiate the above implicitly we get,

$\displaystyle \frac{1}{y} y^{ \prime } = 1$

Which follows

$\displaystyle y^{ \prime} = y$

Notice how $\displaystyle y^{ \prime} = y = f(x)$

This is okay!

$\displaystyle y^{ \prime} = y = e^{x+5}$

We are allowed to have our derivative as a function of our original function, there is no problem.

Knowing this can you now compute your question?