Home > Calculus > Quotient Rule

How do you differentiate #(t^2+2)/(6t-3)^7#?

Answer 1

Isaiah Allen

We must use the quotient rule, which states that

Be #y=f(x)/g(x)#, then #(dy)/(dx)=(f'(x)g(x)-f(x)g'(x))/(g(x))^2#

Now, before starting, we can acknowledge all our four needed functions:

#f'(t)=2t#

#g(t)=(6t-3)^7#

Thus,

#(dy)/(dt)=(2t(6t-3)^7-(t^2+2)(42(6t-3)^6))/(6t-3)^14=#

#=(2tcancel((6t-3)^7))/(6t-3)^(cancel(14)7)-((t^2+2)*42cancel((6t-3)^6))/(6t-3)^(cancel(14)8#

Considering #(6t-3)^8# our l.c.d.:

#(2t(6t-3)-42(t^2+2))/(6t-3)^8=(12t^2-6t-42t^2-84)/(6t-3)^8=color(Green)((-30t^2-6t-84)/(6t-3)^8)#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Christopher Allers

To differentiate (t^2 + 2) / (6t - 3)^7, you would apply the quotient rule of differentiation. The quotient rule states that if you have a function of the form f(x) / g(x), then its derivative is given by [g(x)f'(x) - f(x)g'(x)] / [g(x)]^2. Using this rule:

Find the derivative of the numerator: f'(x) = 2t.
Find the derivative of the denominator: g'(x) = 6.
Substitute these derivatives into the quotient rule formula: [6t(6t - 3)^7 - (t^2 + 2)(6)] / (6t - 3)^14.

This simplifies to:

[12t(6t - 3)^7 - 6(t^2 + 2)] / (6t - 3)^14.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.