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# Calculus help

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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.

May 12, 2018

#1
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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]

May 12, 2018
#2
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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.

May 12, 2018
#3
+110191
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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.

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

Now you can check it and simplify it yoursself.

If you have specific problems then let us know :)

May 12, 2018