(0.6) with a sqare root of 2
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You have y as an implicit function of x:
sin(xy) – x = 0
Use implicit differentiation. As y is a function of x, then you must apply the chain rule there:
d d
—— [ sin(xy) – x ] = —— (0)
dx dx
d d d
—— [ sin(xy) ] – —— (x) = —— (0)
dx dx dx
d
—— [ sin(xy) ] – 1 = 0
dx
d
—— [ sin(xy) ] = 1
dx
d
cos(xy) · —— (xy) = 1
dx
Now, apply the product rule for that last derivative:
![\mathsf{cos(xy)\cdot \left[\dfrac{d}{dx}(x)\cdot y+x\cdot \dfrac{dy}{dx}\right]=1}\\\\\\ \mathsf{cos(xy)\cdot \left[1\cdot y+x\cdot \dfrac{dy}{dx}\right]=1}\\\\\\ \mathsf{y\,cos(xy)+x\,cos(xy)\cdot \dfrac{dy}{dx}=1}](https://tex.z-dn.net/?f=%5Cmathsf%7Bcos%28xy%29%5Ccdot%20%5Cleft%5B%5Cdfrac%7Bd%7D%7Bdx%7D%28x%29%5Ccdot%20y%2Bx%5Ccdot%20%5Cdfrac%7Bdy%7D%7Bdx%7D%5Cright%5D%3D1%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7Bcos%28xy%29%5Ccdot%20%5Cleft%5B1%5Ccdot%20y%2Bx%5Ccdot%20%5Cdfrac%7Bdy%7D%7Bdx%7D%5Cright%5D%3D1%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7By%5C%2Ccos%28xy%29%2Bx%5C%2Ccos%28xy%29%5Ccdot%20%5Cdfrac%7Bdy%7D%7Bdx%7D%3D1%7D)
dy
Isolate —— :
dx
dy
x cos(xy) · —— = 1 – y cos(xy)
dx
Assuming x cos(xy) ≠ 0,
dy 1 – y cos(xy)
—— = ———————— <——— this is the answer.
dx x cos (xy)
I hope this helps. =)
——————————
You have y as an implicit function of x:
sin(xy) – x = 0
Use implicit differentiation. As y is a function of x, then you must apply the chain rule there:
d d
—— [ sin(xy) – x ] = —— (0)
dx dx
d d d
—— [ sin(xy) ] – —— (x) = —— (0)
dx dx dx
d
—— [ sin(xy) ] – 1 = 0
dx
d
—— [ sin(xy) ] = 1
dx
d
cos(xy) · —— (xy) = 1
dx
Now, apply the product rule for that last derivative:
dy
Isolate —— :
dx
dy
x cos(xy) · —— = 1 – y cos(xy)
dx
Assuming x cos(xy) ≠ 0,
dy 1 – y cos(xy)
—— = ———————— <——— this is the answer.
dx x cos (xy)
I hope this helps. =)