How do you find the derivative of #r(x)= (0.3x-4.9x^-1)^0.5#?

1 Answer
Dec 26, 2017

Answer:

#(0.3+4.9/x^2)/(2sqrt(0.3x-4.9/x))#

Explanation:

Let #u=0.3x-4.9x^-1#

#sqrt(u)=sqrt(0.3x-4.9x^-1)#

Using the chain rule,

Differentiate #sqrt(u)# (the function) into #1/(2sqrt(u)# and #0.3x-4.9x^-1# (the inside) into #0.3+4.9/x^2#.

Multiply the two derivatives together to get #1/(2sqrt(u)##(0.3+4.9/x^2)#.

Now, substitute #sqrt(u)=sqrt(0.3x-4.9x^-1)# for #1/(2sqrt(u)#.

Your final answer should be

#1/(2sqrt(0.3x-4.9x^-1)##(0.3+4.9/x^2)# which simplifies into:

#(0.3+4.9/x^2)/(2sqrt(0.3x-4.9/x))#