Find the derivative of the functions

[fr33pl4gu3]

The question is as below:

g(x) = (2x²+x+1) / (x²+2x+1)

It says find the derivative of the functions, I don't get it what does it meant by that, derivative supposed to be dy/dx kind of things right, but all the tutorials is so simple, but how do you do this question??

 

[HallsofIvy]

The question is as below:

g(x) = (2x²+x+1) / (x²+2x+1)

It says find the derivative of the functions, I don't get it what does it meant by that, derivative supposed to be dy/dx kind of things right, but all the tutorials is so simple, but how do you do this question??

First, I am going to move this. Just because it is about a derivative, does not mean it has anything to do with differential equations!

Yes, the derivative is dy/dx or, in this case because the function is called "g", dg/dx.

Since g(x) is defined as a fraction, do you know the "quotient rule"?
\frac{d \frac{f}{g}}{dx}= \frac{\frac{df}{dx}g- f\frac{dg}{dx}}{g^2}

 

[fr33pl4gu3]

No, but i managed to find 2 sites about the question, and i got the final answer in below, i wonder if this is right??

A: (4x+1) / (2x+2)

 

[HallsofIvy]

No, but i managed to find 2 sites about the question, and i got the final answer in below, i wonder if this is right??

A: (4x+1) / (2x+2)

No, it appears that you have just differentiated the numerator and denominator separately and ignored the formula I gave.
\frac{d\frac{f}{g}}{dx}\ne \frac{\frac{df}{dx}}{\frac{dg}{dx}}

Your function is g(x) = (2x²+x+1) / (x²+2x+1) which is of the form f(x)/h(x) with f(x)= 2x²+ x+ 1 and h(x)= x²+ 2x+ 1. Then f'(x)= 4x+ 1 and h'(x)= 2x+ 2. The formula I gave
\frac{d\frac{f}{h}}{dx}= \frac{\frac{df}{dx}g- f\frac{dg}{dx}}{g^2}
then gives
\frac{dg}{dx}= \frac{(4x+1)(x^2+ 2x+1)- (2x^2+ x+ 1)(2x+2)}{(x^2+ 2x+ 1)^2}
= \frac{3x^2+ 2x-1}{(x^2+ 2x+ 1)^2}
which, I believe does not simplify any more.

 

[NoMoreExams]

If you are not familiar with the quotient rule (which you should change immediately), you might be familiar with the product rule. In which case rewrite

g(x) \, = \, \frac{2x^{2} \, + \, x \, + \, 1}{x^{2} \, + \, 2x \, + \, 1}

as

g(x) \, = \, \left( 2x^{2} \, + \, x \, + \, 1 \right) \left(x^{2} \, + \, 2x \, + \, 1 \right)^{-1}

 

[fr33pl4gu3]

New question,

find the derivative of the function,

f(x) = sin²x(3x²-2)⁵=cos²x(3x²-2)⁵ + sin²x(30x-10)⁴

Can i simplify until the answer: (3x²-2)⁵ + (30x-10)⁴

 

[HallsofIvy]

I have no idea what you are doing! You are simply handing us purported "answers"- they are in fact not even close to correct- without showing any work.

Do you know the "chain rule"?
Do you know what the derivative of sin x is?

 

[fr33pl4gu3]

The derivative of the sin x is cos x, right??

 

[NoMoreExams]

The derivative of the sin x is cos x, right??

Yes now do you know the deriviative of f(g(x))?

 

[fr33pl4gu3]

how is product rule different from chain rule??

 

[rootX]

how is product rule different from chain rule??

In way that product rule is for

f(x)*g(x)

so d/dx [f(x)*g(x)] = f'*g + f*g'
and chain rule is for

f[g(x)] = g'(x)*f'[g(x)]

 

[Feldoh]

Chain rule: \frac{d}{dx}[f(g(x))] = g'(x)*f'(g(x))

Product rule: \frac{d}{dx}[f(x)*g(x)] = f'(x)g(x)+g'(x)f(x)

Can you see the differences? It sounds like your trying to teach yourself calculus starting with derivative shortcuts, that's a bad idea. Try looking in the tutorials section for video lectures/e-books instead of reading tutorials on shortcuts.

Edit: Damn beat to it XD

 

[NoMoreExams]

I think the first step is understanding the difference betweeen function multiplication and function composition. Do you know that difference from your algebra class?

 

[fr33pl4gu3]

so the answer should be something like this:

(30x - 10)⁴ * cos²(3x²-2)⁵

 

[rootX]

so the answer should be something like this:

(30x - 10)⁴ * cos²(3x²-2)⁵

>> d/dx '(sin(x))^2*(3*x^2-2)^5'

ans=

It's not

It's 2*sin(x)*(3*x^2-2)^5*cos(x)+30*sin(x)^2*(3*x^2-2)^4*x

This question seems to be above your level right now, try some simpler questions first for good understanding of diff rules and then try this one ...

 

[fr33pl4gu3]

isn't it f(x) = sin²x and g(x) = (3x²-2)⁵?? and how is it possible that it evolve until that stage and how does cos x + 30 came from??

 

[rootX]

Yes, so
it looks like
f[p(x)]*g[q(x)]

Hint: p(x) = sin(x) and f[p(x)] = p(x)^2
similarly q(x) = 3x^2-2 and g(x) = q(x)^5

therefore
d/dx f[p(x)]*g[q(x)] = ??

Use chain rule and product rule together

 

[fr33pl4gu3]

can anyone show me the whole solution to this question, this is my assignment:

f(x) = sin²x(3x²-2)⁵

Please and thank you.

 

[Diffy]

No, I certainly hope no one just shows you the answer, after all it is YOUR assignment. RootX's response should be enough to help you figure it out. What about this derivative is still confusing you? Give it a shot yourself.

 

[fr33pl4gu3]

Is it:
f[p(x)]*g[q(x)]

or

f'[p(x)]*g'[q(x)]

 

[NoMoreExams]

can anyone show me the whole solution to this question, this is my assignment:

f(x) = sin²x(3x²-2)⁵

Please and thank you.

It's not clear what the argument of sine is in your expression, you should use parantheses beter. You should also read up on product and composition of functions, once you've mastered that, read up on the product and chain rule and see if you can get it then. Remember this place is for one to get advice and hints not get your homework solved for you.

 

[fr33pl4gu3]

http://www.math.hmc.edu/calculus/tutorials/chainrule/

In this website, got tutorials about the chain rule but it seems kinda different from yours, i wonder if this can be trusted??

 

[Feldoh]

No, actually that is correct, the chain rule can be written in a multitude of ways.

It might help to re-write the problem like this.

h(x)=sin(x)
j(x)=h(x)^2
g(x)=3x^2-2
k(x)=g(x)^5

So f(x)=j(x)*k(x)