Question

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

Answer 1

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))`