Question

This is Calculus 1 Problem! MUST SHOW ALL THE JUSTIFICATION!!!Find the derivative for this function.

Answer 1

`f^(\prime)(x)=(√(x)(2x-9))/(2(x-3)^(3/2))`

Step-by-step explanation

Before finding the derivative of this function, we can simplify it. This will also simplify the process of derivation.

We include x inside the square root if we square it, and the same for the denominator (x - 3),

`f(x)=(x√(x(x-3)))/(x-3)=\sqrt{(x^2\cdot x(x-3))/((x-3)^2)}=\sqrt{(x^3)/(x-3)}`

Now we apply the chain rule,

`h^(\prime)(g(x))=h^(\prime)(g)\cdot g^(\prime)(x)`

In this case, h(g) is the function f and g(x) is the expression inside the square root,

`f(g)=√(g);g(x)=(x^3)/(x-3)`

Let's find the derivative of f(g). Remember that the square root can be written as a power with exponent 1/2,

`f(g)=g^(1/2)\Rightarrow f^(\prime)(g)=(1)/(2)g^(1/2-1)=(1)/(2)g^(-1/2)=(1)/(2√(g))`

Then, find the derivative of g(x). To do so, we have to apply the quotient rule,

`\left((u(x))/(v(x))\right)^(\prime)=(u^(\prime)(x)\cdot v(x)-u(x)\cdot v^(\prime)(x))/(v^2(x))`

In this case, u(x) = x³ and v(x) = x - 3. Let's find the derivative of each and the square of v(x),

`u^(\prime)(x)=3x^(3-1)=3x^2`
`v^(\prime)(x)=x^(1-1)-0=1`
`v^2(x)=(x-3)^2`

So, the derivative of g(x) is,

`g^(\prime)(x)=(3x^2(x-3)-x^3)/((x-3)^2)=(3x^3-9x^2-x^3)/((x-3)^2)=(2x^3-9x^2)/((x-3)^2)`

Finally, plug in g'(x) and f'(g(x)) in the chain rule expression we found above,

`f^(\prime)(x)=\frac{1}{2\sqrt{(x^3)/(x-3)}}\cdot(2x^3-9x^2)/((x-3)^2)`

This expression can be simplified. First, distribute the square root of the first factor and rewrite it using fractional exponents,

`f^(\prime)(x)=((x-3)^(1/2))/(2x^(3/2))\cdot(2x^(3)-9x^(2))/((x-3)^(2))`

Simplify the exponents of the factor (x - 3),

`f^(\prime)(x)=((x-3)^(1/2-2))/(2x^(3/2))\cdot2x^3-9x^2=((x-3)^(-3/2))/(2x^(3/2))\cdot2x^3-9x^2=(2x^3-9x^2)/(2x^(3/2)(x-3)^(3/2))`

We can also take x² as a common factor in the numerator to simplify that with the denominator,

`f^(\prime)(x)=(x^2(2x^-9))/(2x^(3/2)(x-3)^(3/2))=(x^(2-3/2)(2x^-9))/(2(x-3)^(3/2))=(x^(1/2)(2x-9))/(2(x-3)^(3/2))`

Hence, the derivative of the function is,

`f^(\prime)(x)=(√(x)(2x-9))/(2(x-3)^(3/2))`