# Implicit Differentiation

#### 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?

#### undefined

MHF Hall of Honor
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.

#### AllanCuz

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

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

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?