gwiyomi17 Posted November 7, 2013 Share Posted November 7, 2013 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 Link to comment Share on other sites More sharing options...
studiot Posted November 7, 2013 Share Posted November 7, 2013 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. Link to comment Share on other sites More sharing options...
gwiyomi17 Posted November 7, 2013 Author Share Posted November 7, 2013 (edited) 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 Link to comment Share on other sites More sharing options...
imatfaal Posted November 8, 2013 Share Posted November 8, 2013 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 1 Link to comment Share on other sites More sharing options...
studiot Posted November 8, 2013 Share Posted November 8, 2013 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] 1 Link to comment Share on other sites More sharing options...
imatfaal Posted November 8, 2013 Share Posted November 8, 2013 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 Link to comment Share on other sites More sharing options...
gwiyomi17 Posted November 8, 2013 Author Share Posted November 8, 2013 oh ok, thanks again:) Link to comment Share on other sites More sharing options...
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