Question

if sin(xy) = x, then dy/dx =?

Answer 1

implicit differentiation

\[\frac{d}{dx}(sinxy) = \frac{d}{dx}(x)\]
chain rule:
\[\frac{d}{dx}(xy) \times (cosxy) = 1\]
\[(y + x\frac{dy}{dx}) \times (cosxy) = 1\]

do you see how that works or is it confusing you?

Answer 2

i'm a LITTLE confused..isn't supposed to be sinx * dy/dx + y * something? or am i wrong?

Answer 3

:) to differentiate sin(xy)

the chain rule tells us to differentiate the "outside" function with respect to the "inside"
then differentiate the inside with respect to x and then multiply these two results together
so the outside is sin(xy)

inside is xy

d(sin(xy))/dx = (d(sin(xy))/d(xy))(d(xy)/dx)

if that mess of brackets just made it more confusing im sorry

another way to think about it is:

\[\frac{dv}{dx} = \frac{dv}{du} \times \frac{du}{dx}\]

so in our case v = sin(u)
u =xy

substituting in the above formula gets us

\[\frac{d(sinxy)}{dx} = \frac{dsinu}{du} \times \frac{du}{dx}\]

now we just have to find du/dx , where u = xy

Answer 4

OHHHHHH that actually made way more sense!
thank you so much :)

Answer 5

:D yay thats a relief, i thought i was being really unclear