Differentiation by the Product Rule

[Timiop2008]

Hi
I have been asked to solve the following 2 questions using 'rules' to find the derivative.
(The Product Rule, The Chain Rule or The Quotient Rule.) but I can't remember what these rules are or how they are used

Q1:
Find the equation of the tangent to each of the following curves at the given point:
i) y=x³lnx when x=1
ii) y=x(√3x+1) when x=5
iii) y=x³e^-2x when x=0


I already have the answers but cannot figure out the working out to get to them.
Any help would be very appreciated.
ANSWERS:
i) y=x-1
ii) 8y=47x-75
iii) y=0

 

[CompuChip]

How about you first look up those rules then?
Can you at least remember for which kind of functions they were used? For example, when do you use the product rule?

 

[Timiop2008]

OK

Using the product rule, so far, I have:
i)
y= x³lnx
y= x³.lnx
dy/dx= 3x².lnx + 1/x.x³
dy/dx= 3x²lnx + (x^-1.x³)
dy/dx= 3x²lnx+x²
dy/dx= 4x²lnx

ii)
y= x(3x+1)^1/2
dy/dx= 1.3x+1^1/2+x.3/2x^-1/2
dy/dx= 3x+1^1/2+(3/2x²)^-1/2
dy/dx= 4.5x³+1

iii)
y= x³.e^-2x
dy/dx= 3x².e^-2x+x³.-2e^-2x
dy/dx= 3x²e^-2x+x³-2e^-2x
dy/dx= 3x²-e^-2x+x³

 

[CompuChip]

OK

Using the product rule, so far, I have:
i)
y= x³lnx
y= x³.lnx
dy/dx= 3x².lnx + 1/x.x³
dy/dx= 3x²lnx + (x^-1.x³)
dy/dx= 3x²lnx+x²

So far, you are correct. However, the last step to 4x²lnx is wrong. In the first term you have something with ln(x), you cannot simply add that to a term without ln(x). If you let a = x² and b = ln(x), you are saying that 3 a b + a = 4 a b, which I hope you can see it not true.

iii)
y= x³.e^-2x
dy/dx= 3x².e^-2x+x³.-2e^-2x

Again, you did the differentiation correctly, but you made an error in simplifying.
On the last line I quoted you have x³.-2e^-2x, and on the next line you have removed the dot and written x³-2e^-2x. However, those are NOT the same. In the first case, you mean: take x³ and multiply it by a multiple of e^-2x. In the second case, you mean: take x³ and subtract from that a multiple of e^-2x. The common way of removing the explicit multiplication dot, is writing the numerical factor at the front:
x³-2e^-2x = -2 x³ e^-2x
either of which means: multiply all of minus 2, x³ and e^-2x together.

dy/dx= 3x²e^-2x+x³-2e^-2x
dy/dx= 3x²-e^-2x+x³

[/QUOTE]
Also here you are making the same mistake as in i): e^-2x and x²e are different things, you cannot just add them together.

ii)
y= x(3x+1)^1/2
dy/dx= 1.3x+1^1/2+x.3/2x^-1/2
dy/dx= 3x+1^1/2+(3/2x²)^-1/2
dy/dx= 4.5x³+1

I cannot really see what you are doing there. By the product rule,
dy/dx = 1 . (3x + 1)^1/2 + x d/dx( (3x+1)^1/2 )
and you need the chain rule to evaluate the derivative of (3x+1)^1/2.

In other words, your differentiation techniques are probably better than you think, but you really need to brush up on your algebra. Be careful when working with brackets (you cannot write (3x + 1)^1/2 = (3x)^1/2 + 1^1/2, for example) and watch out that when you combine terms, they only differ by a number. For example, you can add 2x to 3x to get 5x, but you can't add 2x to 5x² to get 7x², or something like that.

 

[Timiop2008]

Thank You
But how does this link to the answers given of:
i) y=x-1
ii) 8y=47x-75
iii) y=0
when the answers we are discussing are so different
ie i) 3x²lnx
iii) 3x²-e^-2x+x³

 

[CompuChip]

I have no idea what those answers have to do with it.
Especially since you are given values of x at which to evaluate the derivatives, so you should just get a number in the end.

Are you sure they are the answers to the question you are asking?

 

[HallsofIvy]

CompuChip, the question asked was to "find the equation of the tangent to each of the following curves at the given point", not just to find the derivative (although that was how the title to the thread was given).

Timiop2008, these question assume that you know that the slope of the tangent line is the derivative evaluated at the given point and that you know how to find the equation of a line given the slope and a point on the line. Do you know those things?