derivative: f(x) = (x^2)sqrt(5 - x^2), f(x) = (x^3)sin^2(5x)

[Guest]

Find f' for:

1) f(x)=(x^2)sqrt(5-x^2)

2) f(x)=(x^3)sin^2(5x)

For (1), I tried to take the derivative of x^2 , sqrt of 5-x^2 and 5-x^2 then multiplied those together. But I keep getting a weird answer. I can't get the right answer for (2) either.

The answer in the book for (1) is x(10-3x^2)/(sqrt(5-x^2))

And the answer for (2) is 10(x^3)sin(5x)cos(5x)+3(x^2)sin^2(5x)

I don't really get why there is a plus in the answer for two. :?: :?: :?:

Thanks for your help.

 

[galactus]

I assume you know you need the product rule and a little chain rule.

\(\displaystyle \frac{d}{dx}[x^{2}\sqrt{5-x^{2}}]\)


\(\displaystyle x^{2}(\frac{1}{2})(\sqrt{5-x^{2}})^{\frac{-1}{2}}(2x)+2x\sqrt{5-x^{2}}\)

\(\displaystyle =\frac{-x^{3}}{\sqrt{5-x^{2}}}+2x\sqrt{5-x^{2}}\)

Now, this simplifies to what they have i the back of your book. You didn't by chance get this far and think you had the wrong answer, did you?. It's the same thing, only another form.

If you multiply by the LCD and factor, you get:

\(\displaystyle \frac{-x^{3}+2x(5-x^{2})}{\sqrt{5-x^{2}}}\)

\(\displaystyle =\frac{x(10-3x^{2})}{\sqrt{5-x^{2}}}\)


The 2nd one is done the same way.

\(\displaystyle \frac{d}{dx}[x^{3}sin^{2}(5x)]\)

\(\displaystyle 10x^{3}cos(5x)sin(5x)+3x^{2}sin^{2}(5x)\)

Factor if you want:

\(\displaystyle x^{2}sin(5x)(10xcos(5x)+3sin(5x))\)

 

[soroban]

Re: derivative

Hello, Bryan!

Find \(\displaystyle f'(x)\)

\(\displaystyle 1)\;f(x)\:=\: x^2\cdot\sqrt{5\,-\,x^2}\)

\(\displaystyle 2)\:f(x)\:=\:x^3\cdot\sin^2(5x)\)

for number one, i tried to take the derivative of x^2 , sqrt of 5-x^2 and 5-x^2
then multiplied those together \(\displaystyle \;\;\) . . . what?

Answers: \(\displaystyle \L\:\begin{array}{cc} 1)\;\frac{x(10\,-\,3x^2)}{\sqrt{5\,-\,x^2}} \\ 2)\;10x^3\cdot\sin(5x)\cdot\cos(5x)\,+\,3x^2\cdot\sin^2(5x)\end{array}\)

i dont really get why there is a plus in the answer for #2.
It seems you don't know the Product Rule . . . bummer!

Cody (Galactus) did an excellent job . . . as always.
Let me baby-step through #1 . . .

\(\displaystyle 1)\;y\;=\;\underbrace{x^3}\cdot\underbrace{\left(5\,-\,x^2)^{\frac{1}{2}}}\)
. . . . . \(\displaystyle f(x)\;\;g(x)\)

Then: \(\displaystyle \,y'\;=\;\underbrace{x^3}\cdot\underbrace{\frac{1}{2}\cdot\left(5 - x^2\right)^{-\frac{1}{2}}(-2x)}\:+\:\underbrace{2x}\cdot\underbrace{\sqrt{5\,-\,x^2}}\)
. . . . . . . . .\(\displaystyle f(x)\;\;\;\;\;g'(x)\;\;\;\;\;\;\;\;f'(x)\;\:g(x)\)

We have: \(\displaystyle \:y'\;=\;-x^3(5\,=\,x^2)^{-\frac{1}{2}}\,+\,2x(5\,-\,x^2)^{\frac{1}{2}}\)


Multiply top and bottom by \(\displaystyle (5\,-\,x^2)^{\frac{1}{2}}\)

. . \(\displaystyle \L y'\;= \;\frac{(5\,-\,x^2)^{\frac{1}{2}}}{(5\,-\,x^2)^{\frac{1}{2}}}\,\cdot\,\frac{-x^3(5\,-\,x^2)^{-\frac{1}{2}} \,+\,2x(5\,-\,x^2)^{\frac{1}{2}}}{1} \;=\;\frac{-x^3\,+\,2x(5\,-\,x^2)}{(5\,-\,x^2)^{\frac{1}{2}}}\)

. . \(\displaystyle \L\:y'\;=\;\frac{-x^3\,+\,10x\,-\,2x^3}{(5\,-\,x^2)^{\frac{1}{2}}} \;= \;\frac{10x\,-\,3x^3}{(5\,-\,x^2)^{\frac{1}{2}}} \;= \;\frac{x(10\,-\,3x^2)}{\sqrt{5\,-\,x^2}}\)

 

[Guest]

Thanks for the great replies.
I probably forgot to multiply it out first then find the derivative.
I am studying cal AB in summer and taking BC in fall so things seem to come very fast and I dont have much time to review.