What is the derivative of #f(x) = (x)/(sqrt(7-3x))#?

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Answer 1 · 2015-06-07T16:20:13

We can rewrite the denominator using a property of exponentials that states #a^(m/n)=root(n)(a^m)#:

#f(x)=x/(7-3x)^(1/2)#

Also, another law of exponentials states that #a^-n=1/a^n#. Thus,

#f(x)=x(7-3x)^(-1/2)#

Now, we can use the product rule, where #g(x)=x# and #h(x)=(7-3x)^(-1/2)#, following the rule statement:

#y=g(x)/h(x)# then #y'=g'(x)h(x)+g(x)h'(x)#

Thus, we need to find #g'(x)# and #h'(x)# and, then, proceed to the chain rule.

#g'(x)=1#
To find #h'(x)#, we need chain rule, which states that #(dy)/(dx)=(dy)/(du)(du)/(dx)#. Renaming #u=7-3x#, we have #h(x)=u^(-1/2)#, which we can derivate using power rule:

#h'(x)=-1/2u^(-3/2)(2x)=-(cancel(2)x)/(cancel(2)(7-3x)^(3/2))#

Now, proceeding to the product rule:

#(dy)/(dx)=1/(7-3x)^(1/2)-(x^2)/(7-3x)^(3/2)#

An exponential law states that #a^n*a^m=a^(n+m)#, thus, we can sum the expressions having #(7-3x)^(3/2)# as our l.c.d.

#(dy)/(dx)=color(green)((7-3x+x^2)/(7-3x)^(3/2))#