behemuthxero said: »
Example)
d/dx[(x^2)/(sin(x))]
Recall:
d/dx{ln(f(x))] = (f'(x))/(f(x))
Let f(x) = (x^2)/(sin(x))
Then,
d/dx[ln(f(x)] = d/dx[ln((x^2)/(sin(x))] =
d/dx[ln((x^2)) - ln((sin(x))]=
d/dx[ln((x^2))] - d/dx[ln((sin(x))].
Using the above for the derivative of a logarithm we have:
d/dx[ln((x^2))] = (2x)/(x^2)
and
d/dx[ln((sin(x))] = (cos(x))/(sin(x)).
Thus,
d/dx[ln((x^2)/(sin(x))] =
(f'(x))/(f(x)) = ((2x)/(x^2)) - ((cos(x))/(sin(x))).
Multiply through by f(x) yields:
f'(x) = ((x^2)/(sin(x))) * ((2x)/(x^2)) - ((cos(x))/(sin(x))).
Simplify:
f'(x) = ((2x)/(sin(x))) - ((x^2)(cos(x))/(sin^2(x)))=
((2x)(sin(x))/(sin^2(x))) - ((x^2)(cos(x))/(sin^2(x)))=
((2x)(sin(x))-(x^2)(cos(x)))/(sin^2(x)).
From which it can easily be seen for p(x)/q(x)
d/dx[p(x)/q(x)] = ((p'(x)*q(x))-(p(x)*q'(x)))/((q(x)^2)).
d/dx[(x^2)/(sin(x))]
Recall:
d/dx{ln(f(x))] = (f'(x))/(f(x))
Let f(x) = (x^2)/(sin(x))
Then,
d/dx[ln(f(x)] = d/dx[ln((x^2)/(sin(x))] =
d/dx[ln((x^2)) - ln((sin(x))]=
d/dx[ln((x^2))] - d/dx[ln((sin(x))].
Using the above for the derivative of a logarithm we have:
d/dx[ln((x^2))] = (2x)/(x^2)
and
d/dx[ln((sin(x))] = (cos(x))/(sin(x)).
Thus,
d/dx[ln((x^2)/(sin(x))] =
(f'(x))/(f(x)) = ((2x)/(x^2)) - ((cos(x))/(sin(x))).
Multiply through by f(x) yields:
f'(x) = ((x^2)/(sin(x))) * ((2x)/(x^2)) - ((cos(x))/(sin(x))).
Simplify:
f'(x) = ((2x)/(sin(x))) - ((x^2)(cos(x))/(sin^2(x)))=
((2x)(sin(x))/(sin^2(x))) - ((x^2)(cos(x))/(sin^2(x)))=
((2x)(sin(x))-(x^2)(cos(x)))/(sin^2(x)).
From which it can easily be seen for p(x)/q(x)
d/dx[p(x)/q(x)] = ((p'(x)*q(x))-(p(x)*q'(x)))/((q(x)^2)).
There was a time where i could follow that fairly easily... Now i think it looks remarkable similar to Chinese script /grin :(