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

Question

Elijah Abram · Accepted Answer

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

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.

Adam Adkins · Answer

undefined To differentiate $ y = \frac{1}{t^2 + 3x - 1} $ using the chain rule, you first rewrite the function as $ y = (t^2 + 3x - 1)^{-1} $. Then, you differentiate with respect to $ t $ treating $ t^2 + 3x - 1 $ as the inner function and $ u = t^2 + 3x - 1 $ as the outer function. The derivative of $ u $ with respect to $ t $ is $ \frac{du}{dt} = 2t $. Then, you apply the chain rule, which states that $ \frac{dy}{dt} = \frac{dy}{du} \cdot \frac{du}{dt} $. So, $ \frac{dy}{du} = -1 \cdot (t^2 + 3x - 1)^{-2} $ and $ \frac{dy}{dt} = -\frac{2t}{(t^2 + 3x - 1)^2} $.