Question

differentiate implicitly..
x sin ( x square times y cube )=y

Answer 1

\[x \sin (x^{2}y{2} )=y\]

Answer 2

x sin ( x^2 * y^3 ) = y ?

Answer 3

implicit is just a fancy name for; do the same thing as explicit just dont favor any variables

Answer 4

how

Answer 5

differentiate implicitly

Answer 6

you have a product rule; a chain rule; and a product rule inside of the chain rule :)

Answer 7

x [sin(x^2 y^2)]' [x^2 y^2]' + x' sin(x^2 y^2) = y'

Answer 8

let's see if i got it:
\[y^2\cos (x^2y^3)+2x \cos (x^2y^3).(dy/dx)=y \ln x.(dy/dx)-yx^-2\]

Answer 9

y^3 ... but still the same process

Answer 10

Just differenciate both sides of the equation and solve for y':

\begin{eqnarray*}&&\frac{d}{dx} \left(x \sin(x^2 y^3\right) = \frac{d}{dx}y \\ &\Rightarrow& \sin\left(x^2 y^3\right) + x \cos\left(x^2 y^3\right) \left(2x y^3 + 3x^2y^2y'\right) = y' \\ &\Rightarrow& y' = \frac{\sin\left(x^2y^3\right) + 2x^2y^3\cos\left(x^2y^3\right)}{1-3x^3y^2\cos\left(x^2y^3\right)}.\end{eqnarray*}

Answer 11

\[y = x \sin(x ^{2}y ^{3})\]
\[y' = x \sin(x ^{2}y ^{3}) + x(2xy ^{3} + 3x ^{2}y ^{2}y')\cos(x ^{2}y ^{3})

Answer 12

i hope God will repay your kindness