8x^3+3y^3=11xy
I need to find the derivative
And I would really like if someone helped explain the problem please
with respect to x ??
Yes
ok. are you familiar with the power rule ?? \[8x^3+3y^3=11xy\] just take the derivative of each term at left side and at right side apply the power rule.
Um can you explain the power rule please
hmm okay \[y = x^n \] \[y'=n x^{n-1} \] power rule: multiply the coefficient by the exponent and then subtract one from the exponent.
\[y= 2x^5\] for example the derivative of this function would be \[\frac{dy}{dx}= 2\cdot 5 x^{5-1} \frac{dx}{dx}\] \[\frac{dy}{dx}= 10 x^{4} \]
Ok thanks I got it now
alright so what would be the derivative of 8x^3 and 3y^3 ??
On the left 24x^2+9y^2=11y
+11xdy with respect to the X ?
^2 means it is supposed to be squared
\[8x^3+9y^3=11xy\] ignore the right side for a sec don't forget the notation \[24 x^2 \frac{dx}{dx} +9y^3=11xy\] 24x^2 dx/dx you're taking the derivative of x with respect x so it would be dx/dx=1 \[24 x^2 (1)+9y^2 \frac{dy}{dx}=11xy\] we are taking the derivative of y with respect to x so it should be 9y^2 dy/dx
notation dy/dx or dx/dx
no. for the right side we have to apply the product rule because it's a product of two variables 11x times y the product rule:\[f(x) * g(x) = f'(x) *g(x) + f(x) *g'(x)\] Derivative of 1st function times the 2nd function + 1st function times the derivative of 2nd function
So if I apply the product rule correctly I should get it ?
you have to simplify at the end. we are solving for dy/dx
For me it's not so much the but the process but I think I got it I just need to remember the steps and to simplify and I have to go now :(
THE answer*
Thanks for guiding me I think I got it now :)
yw. and yes you should work on a lot of questions then you will end up memorizing these rules :)
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