We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website.
Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see
cookie policy and privacy policy.
DECLINE COOKIES

Find the derivative of the function h(t)=(t^4-1)(t^3+1)^4 using either quotient rule, product rule, addition rule, subtraction rule, and/or chain rule. Please show step by step instructions. Thanks.

gibsonj338 May 12, 2018

#1**+2 **

Find the derivative of the following via implicit differentiation:

d/dt(h(t)) = d/dt((1 + t^3)^4 (-1 + t^4))

The derivative of h(t) is h'(t):

h'(t) = d/dt((1 + t^3)^4 (-1 + t^4))

Use the product rule, d/dt(u v) = v ( du)/( dt) + u ( dv)/( dt), where u = (t^3 + 1)^4 and v = t^4 - 1:

h'(t) = (t^4 - 1) d/dt((1 + t^3)^4) + (t^3 + 1)^4 d/dt(-1 + t^4)

Using the chain rule, d/dt((t^3 + 1)^4) = ( du^4)/( du) ( du)/( dt), where u = t^3 + 1 and d/( du)(u^4) = 4 u^3:

h'(t) = (1 + t^3)^4 (d/dt(-1 + t^4)) + (-1 + t^4) 4 (t^3 + 1)^3 d/dt(1 + t^3)

Differentiate the sum term by term:

h'(t) = (1 + t^3)^4 (d/dt(-1 + t^4)) + d/dt(1) + d/dt(t^3) 4 (1 + t^3)^3 (-1 + t^4)

The derivative of 1 is zero:

h'(t) = (1 + t^3)^4 (d/dt(-1 + t^4)) + 4 (1 + t^3)^3 (-1 + t^4) (d/dt(t^3) + 0)

Simplify the expression:

h'(t) = 4 (1 + t^3)^3 (-1 + t^4) (d/dt(t^3)) + (1 + t^3)^4 (d/dt(-1 + t^4))

Use the power rule, d/dt(t^n) = n t^(n - 1), where n = 3.

d/dt(t^3) = 3 t^2:

h'(t) = (1 + t^3)^4 (d/dt(-1 + t^4)) + 3 t^2 4 (1 + t^3)^3 (-1 + t^4)

Simplify the expression:

h'(t) = 12 t^2 (1 + t^3)^3 (-1 + t^4) + (1 + t^3)^4 (d/dt(-1 + t^4))

Differentiate the sum term by term:

h'(t) = 12 t^2 (1 + t^3)^3 (-1 + t^4) + d/dt(-1) + d/dt(t^4) (1 + t^3)^4

The derivative of -1 is zero:

h'(t) = 12 t^2 (1 + t^3)^3 (-1 + t^4) + (1 + t^3)^4 (d/dt(t^4) + 0)

Simplify the expression:

h'(t) = 12 t^2 (1 + t^3)^3 (-1 + t^4) + (1 + t^3)^4 (d/dt(t^4))

Use the power rule, d/dt(t^n) = n t^(n - 1), where n = 4.

d/dt(t^4) = 4 t^3:

**h'(t) = 12t^2 (1 + t^3)^3 (-1 + t^4) + 4t^3 (1 + t^3)^4 [Courtesy of Mathematica 11 Home Edition]**

Guest May 12, 2018

#2**0 **

Whoops. I forgot to put to the third power on (t^4-1). The equation should be h(t)=(t^4-1)^3(t^3+1)^4. Can you please do it again? I am so sorry.

gibsonj338 May 12, 2018

#3**+1 **

Gibson, you would have been better off just to say thanks to guest , and give a point and then put your 'real' question as another post.

You never had to admit that your first post was an error.

\(h(t)=(t^4-1)^3(t^3+1)^4\\ let \;u=(t^4-1)^3\\ \frac{du}{dt}=3(t^4-1)^2*4t^3\\ \frac{du}{dt}=12t^3(t^4-1)^2\\ let\;v=(t^3+1)^4\\ \frac{dv}{dt}=4(t^3+1)^3*3t^2\\ \frac{dv}{dt}=12t^2(t^3+1)^3\\ \text{now use product rule}\\ If \;y=uv\;\;then\;\;\frac{dy}{dt}=uv'+vu'\\ \frac{dy}{dt}=[(t^4-1)^3][12t^2(t^3+1)^3]+[(t^3+1)^4][12t^3(t^4-1)^2] \)

Now you can check it and simplify it yoursself.

If you have specific problems then let us know :)

Melody May 12, 2018