Differentiation of a product: sqrt(x^3) ln(3x)

[Anthonyk2013]

Is this correct and I yes can I simplify more?

\(\displaystyle 5.\, \sqrt{\strut x^3\, }\, \ln(3x)\)

\(\displaystyle u\, =\, \sqrt{\strut x^3\,}\, =\, x^{\frac{3}{2}}\)

\(\displaystyle \dfrac{du}{dx}\, =\, \dfrac{3}{2}\, x^{\frac{1}{2}}\, =\, \dfrac{3}{2\, \sqrt{\strut x\,}}\)

\(\displaystyle v\, =\, \ln(3x)\)

\(\displaystyle \dfrac{dv}{dx}\, =\, \dfrac{1}{3x}\, \times\, 3\, =\, \dfrac{1}{x}\)

\(\displaystyle \dfrac{dy}{dx}\, =\, \bigg(\, \sqrt{\strut x^3\, }\, \bigg)\, \bigg(\, \dfrac{1}{x}\, \bigg)\, +\, \bigg(\, \ln(3x)\, \bigg)\, \bigg(\, \dfrac{3}{2\, \sqrt{\strut x\,}}\, \bigg)\)

 

[Deleted member 4993]

Is this correct and I yes can I simplify more?

\(\displaystyle 5.\, \sqrt{\strut x^3\, }\, \ln(3x)\)

\(\displaystyle u\, =\, \sqrt{\strut x^3\,}\, =\, x^{\frac{3}{2}}\)

\(\displaystyle \dfrac{du}{dx}\, =\, \dfrac{3}{2}\, x^{\frac{1}{2}}\, =\, \dfrac{3}{2\, \sqrt{\strut x\,}}\)

\(\displaystyle v\, =\, \ln(3x)\)

\(\displaystyle \dfrac{dv}{dx}\, =\, \dfrac{1}{3x}\, \times\, 3\, =\, \dfrac{1}{x}\)

\(\displaystyle \dfrac{dy}{dx}\, =\, \bigg(\, \sqrt{\strut x^3\, }\, \bigg)\, \bigg(\, \dfrac{1}{x}\, \bigg)\, +\, \bigg(\, \ln(3x)\, \bigg)\, \bigg(\, \dfrac{3}{2\, \sqrt{\strut x\,}}\, \bigg)\)

Correct

You can simplify the first term.

 

[Anthonyk2013]

Correct

You can simplify the first term.

Perfect thanks.

 

[Ishuda]

Is this correct and I yes can I simplify more?

\(\displaystyle 5.\, \sqrt{\strut x^3\, }\, \ln(3x)\)

\(\displaystyle u\, =\, \sqrt{\strut x^3\,}\, =\, x^{\frac{3}{2}}\)

\(\displaystyle \dfrac{du}{dx}\, =\, \dfrac{3}{2}\, x^{\frac{1}{2}}\, =\, \dfrac{3}{2\, \sqrt{\strut x\,}}\) <=============

\(\displaystyle v\, =\, \ln(3x)\)

\(\displaystyle \dfrac{dv}{dx}\, =\, \dfrac{1}{3x}\, \times\, 3\, =\, \dfrac{1}{x}\)

\(\displaystyle \dfrac{dy}{dx}\, =\, \bigg(\, \sqrt{\strut x^3\, }\, \bigg)\, \bigg(\, \dfrac{1}{x}\, \bigg)\, +\, \bigg(\, \ln(3x)\, \bigg)\, \bigg(\, \dfrac{3}{2\, \sqrt{\strut x\,}}\, \bigg)\)

Not quite what I get. Am I missing something?
\(\displaystyle \dfrac{3}{2}\, x^{\frac{1}{2}}\, \ne\, \dfrac{3}{2\, \sqrt{\strut x\,}}\)
unless, of course, x=1.

 

[Anthonyk2013]

Not quite what I get. Am I missing something?
\(\displaystyle \dfrac{3}{2}\, x^{\frac{1}{2}}\, \ne\, \dfrac{3}{2\, \sqrt{\strut x\,}}\)
unless, of course, x=1.

So you cat bring the X back under the line using indices? Think I made same mistake before.

 

[Ishuda]

So you cat bring the X back under the line using indices? Think I made same mistake before.

You can if it goes there but
\(\displaystyle \dfrac{3}{2}\, x^{\frac{1}{2}}\, =\, \dfrac{3\, \sqrt{\strut x}}{2}\)
and
\(\displaystyle \dfrac{3}{2}\, x^{-\frac{1}{2}}\, =\, \dfrac{3}{2 \, \sqrt{\strut x}}\)
Notice the sign of the exponent on the x.

 

[Anthonyk2013]

You can if it goes there but
\(\displaystyle \dfrac{3}{2}\, x^{\frac{1}{2}}\, =\, \dfrac{3\, \sqrt{\strut x}}{2}\)
and
\(\displaystyle \dfrac{3}{2}\, x^{-\frac{1}{2}}\, =\, \dfrac{3}{2 \, \sqrt{\strut x}}\)
Notice the sign of the exponent on the x.

Silly errors will cost me. Thanks