Implicit Differentiation
Neil Trivedi
Teacher
Implicit Differentiation
Implicit differentiation is a technique that is useful for when neither nor can be made the subject.
For example, finding for . Although it is possible to rearrange for either or here, it is a big task and not necessary if we just want the gradient function. In this case, using implicit differentiation will allow us to find the gradient in terms of both and instead of just one of them.
We still differentiate each term of our equation with respect to , but every time we differentiate a function of , we treat it like a function of and multiply it by . Here is how it works:
We want to differentiate a Instead, we differentiate with respect to , but we
function of with respect to . cannot just replace with . We need to
balance the equation using the chain rule.
Notice how the ’s cancel out.
Note: We have been using this concept already. When we differentiate , it gives us .
Here, we are differentiating to give and then multiplying it by to leave us with (mind blown moment!)
Example 1:
Differentiate each function with respect to .
a)
Single Step: and are two separate functions so we apply the product rule with the formula .
When it comes to differentiating , remember we differentiate it as normal to give , then multiply by to give .
Differentiating with respect to gives:
b)
Single Step: and are two separate functions so we apply the product rule together with the formula .
Differentiating with respect to gives:
No answer provided.
Example 2:
Find in terms of and in each case.
a)
Step 1: Differentiate each term of the equation individually.
When it comes to differentiating , remember we differentiate it as normal to give , then multiply by to give .
Step 2: Rearrange to obtain .
Dividing both sides by , we are left with:
b)
Step 1: Differentiate each term of the equation individually.
To differentiate , we use the chain rule for exponential functions. The is a multiplier so we write that out in front. The angle is the power, which is . We differentiate the angle to give . Then, the function of stays the same in differentiation as well as its power.
Multiplier Angle stays the same
Differentiated angle
When it comes to differentiating , we differentiate it as normal using the chain rule for exponential functions, then multiply by at the end. In this case, the angle is , which differentiates to . Then, the function of stays the same in differentiation as well as its power.
Angle stays the same
Differentiated angle
Therefore,
Step 2: Rearrange to obtain .
Dividing both sides by ,
No answer provided.
Example 3:
The curve has the equation .
The point , with coordinates lies on .
Find the gradient of at .
Step 1: Differentiate each term of the equation individually.
differentiates to and differentiates to .
For and are two separate functions, so we apply the product rule with the formula, . When it comes to differentiating , we differentiate it as normal to give , then multiply by to get .
Differentiating with respect to gives:
For the right side, we will differentiate using the chain rule for exponential functions. The is a multiplier so we write that out first. The angle of the function is the power, which is in this case, and that differentiates to . Then, the base (which is ) is kept the same as well as the power. Then, we apply to the base.
Multiplier Angle stays the same
Same base Multiply by of the base
Therefore,
Step 2: Find the gradient by substituting the coordinates of .
Note: Students make many silly mistakes here as they try to rearrange for first. When they give you coordinates, substitute first.
We are given the coordinates . We substitute that into to find the gradient of at this point.
Rearranging for , we get:
Therefore, the gradient of at is .
No answer provided.
Challenging Question