How do you use the chain rule to differentiate y=1/(t^2+3x-1)?

1 Answer
Apr 11, 2018

dy/dx=(-3)/(t^2 + 3x - 1)^2 or dy/dt = (-2t)/(t^2 + 3x - 1)^2

Explanation:

You need to decide whether you want to differentiate the expression with respect to t or x i.e whether you want dy/dx or dy/dt but the calculation is very similar in either case.

Let's first change the original expression to make it easier to differentiate:

y = 1 / (t^2 + 3x - 1) =(t^2 + 3x - 1)^-1

Now the chain rule tells us that you will need to differentiate this expression twice. First, what is outside the brackets ("formula")^-1. Second, what is inside the brackets (t^2 + 3x - 1). Note that I use "formula" as a generic term to stand for what is inside the brackets: (t^2 + 3x - 1).

Assume we want to calculate dy/dx.

We start by differentiating the expression outside the brackets to get:

dy/dx = -1*("formula")^-2 = -1/("formula")^2

We then differentiate the what is inside the brackets:

dy/dx(t^2 + 3x - 1)= 3

Now by the chain rule we just multiply these two expressions together to get:

dy/dx = 3 * -1/("formula")^2

Substituting the whole expression in "formula", this becomes:

dy/dx = (-3)/(t^2 + 3x - 1)^2

This is the final answer.

Assume we want to calculate dy/dt the only difference is the second step when we differentiate what is inside the bracket:

dy/dt(t^2 + 3x - 1)= 2t

So that the final step by the chain rule is:

dy/dt = 2t* -1/("formula")^2 = (-2t)/(t^2 + 3x - 1)^2

This is the final answer.