+0  
 
0
49
3
avatar+1837 

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
Sort: 

3+0 Answers

 #1
avatar
+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
avatar+1837 
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
avatar+92463 
+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

11 Online Users

avatar
New Privacy Policy (May 2018)
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  Privacy Policy