Question

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

Answer 1

`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.