Can someone check my answer/solution :)

[gwiyomi17 8]

 

i'm not sure if i'm right,

i know there's another method in solving this but i'm not sure how to do that

[studiot 2108]

 

i'm not sure if i'm right,

 

Sorry you are off beam here.

 

for f(x) = A(x)+B(x)+C(x)

 

take logs

 

ln(A(x)+B(x)+C(x)) is not ln A + lnB + lnC.

 

lnA + lbB +lnC is the log of (A times B times C)

 

So have another go.

[gwiyomi17 8]

oh yeah, i'm totally out of the track,

i need to do it again :/

thanks

am i right now?

 

Edited November 7, 2013 by gwiyomi17

[imatfaal 2481]

I would be more explicit and make it clear where you are using substitution and the chain rule etc. Lumping together stages might sometimes save time but can also mean that the calculation cannot be followed (either by an examiner or by yourself later on) - you have a use of the chain rule, the product rule and a simple derivative taken all as one unexplained stage. The answer and method is as I would do it - but in the end I would include more lines of working

[studiot 2108]

Looks much better

 

A small point

 

For differentiation with respect to x (d.w.r.t.x) we write

[math]\frac{d}{{dx}}(something)[/math]

not

[math]\frac{{dy}}{{dx}}(something)[/math]

 

So your first line should be

[math]\frac{d}{{dx}}({e^{xy}}) - \frac{d}{{dx}}({x^3}) + \frac{d}{{dx}}(3{y^2}) = \frac{d}{{dx}}(1)[/math]

[imatfaal 2481]

Looks much better

 

A small point

 

For differentiation with respect to x (d.w.r.t.x) we write

[math]\frac{d}{{dx}}(something)[/math]

 

not

[math]\frac{{dy}}{{dx}}(something)[/math]

 

So your first line should be

[math]\frac{d}{{dx}}({e^{xy}}) - \frac{d}{{dx}}({x^3}) + \frac{d}{{dx}}(3{y^2}) = \frac{d}{{dx}}(1)[/math]

 

 

 

Agree entirely - more than ever in this case as you are not simply differentiating a y = some polynomial function of x; but are working to get dy/dx as an answer which arises from the two instances of d/dx of y

[gwiyomi17 8]

oh ok, thanks again:)