# Math Help - Implicit Differentiation

1. ## 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.

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

2. 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.

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.

3. 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.

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

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

And

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

If we want to differentiate the above implicitly we get,

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

Which follows

$y^{ \prime} = y$

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

This is okay!

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

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?