Posted by T.Gracken on August 12, 2002 at 10:16:08:
In Reply to: differentiation posted by bana on August 12, 2002 at 07:35:04:
: Hi,
: I've got some more differentiations I don't understand how they got the answers.
: 1) f(x)=(2Inx)(x^3+3), f'(x)=6x^2Inx+2x^2+6/x
: I tried using the log rule where Inx=1/x and the multiplying rule, but I got confused because of all the brackets. could u show me the main line of working please.
I will assume that you mean ln(x) where you have typed Inx.
please realize ln(x) does not equal 1/x. The derivative of [ln(x)] is equal to 1/x.
...some notation for you (that I will use here). If u is a function of x, then Dx[u] represents the derivative of u with respect to x.
now, if f(x) = 2ln(x)*(x3+3), and you want f'(x), start with the product rule.
f'(x) = 2ln(x)* Dx[x3+3] + Dx[2ln(x)]*(x3+3) .......use of product rule
= 2ln(x)*[3x2] + 2(1/x)*(x3+3) ......use power rule and natural log. rule with constant factor rule
= 6x2ln(x) + [2x3+6]/x .....algebra
= 6x2ln(x) + 2x2+6/x .....more algebra
: also,
: 2) y=(5-2x)^1/2
: how do we approach this question? do we multiply (5-2x) by 1/2 and then take it to the power of -1/2?
: thanx for your time.
there is a little more here. you can use the chain rule or the power rule for functions.
the power rule for functions is: Dx[f(x)n] = n*[f(x)n-1]*Dx[f(x)] .....there are some conditions that are placed on f(x) and n but I'm not going to write them here; look them up in your text.
So, for y = (5-2x)1/2,
y' = (1/2)*(5-2x)-1/2*Dx[5-2x]
= (1/2)(5-2x)-1/2(-2)
= you finsh the algebra...
I hope that helps, and I REALLY hope I didn't screw up on the html tags! (if I did I am sure to hear about it!!