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

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Answer 1 · 2017-12-26T08:11:34

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

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))#