How do I differentiate the following equations with roots?

a.) #root(3)(t)/(t-9#
b.) H(u) = (u-#sqrt(u)#) *(u + #sqrt(u)#)

1 Answer
Øko · Jim H
Feb 14, 2018

#f'(t)=(1/3t^(-2/3)-2t^(1/3))/(t-9)^2#
#H'(u)=2u-1#

Explanation:

a)

We want to find the derivative of

#f(t)=root(3)t/(t-9)=t^(1/3)/(t-9)#

Use the Quotient Rule, if #f(t)=(h(t))/g(t)#,

Then #f'(t)=(h'(t)g(t)-h(t)g'(t))/(h(t))^2#

By the Quotient Rule with #h(t)=t^(1/3)# and #g(t)=t-9#
and the Power Rule for differentiation

#f'(t)=((d/dx(t^(1/3)))(t-9)-t^(1/3)(d/dx(t-9)))/(t-9)^2#

#=(1/3t^(-2/3)(t-9)-t^(1/3))/(t-9)^2#

#=(1/3t^(1/3)-3t^(-2/3)-t^(1/3))/(t-9)^2#

#=(-3t^(-2/3)-2/3t^(1/3))/(t-9)^2#

#=(-9t^(-2/3)-2t^(1/3))/(3(t-9)^2)#

#=(-9-2t)/(3(t-9)^2t^(2/3))#

b)

We want to find the derivative of

#H(u)=(u-sqrt(u))(u+sqrt(u))#

We could use the Product Rule

But here it would be easier to rewrite as

#H(u)=(u-sqrt(u))(u+sqrt(u))=u^2-u#

Use the Power Rule , if #f(x)=x^n#

Then #f'(x)=nx^(n-1)#

By the Power Rule

#H'(u)=2u^(2-1)-u^(1-1)#

#=2u-1#

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