anonymous
  • anonymous
I need to find the slope and y-intercept of 5x-2y=3. I think the slope is 5/2 and the y-intercept is -3/2???? Is this correct?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
yes it is .....can we work on a partial diffferential problem?
anonymous
  • anonymous
sorry don't have any problems like that right now.
anonymous
  • anonymous
i do have ...and need someone to workout on it...i want to match my answer with it

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anonymous
  • anonymous
slope and intercept on y axis are correct
anonymous
  • anonymous
whats the second order partial derivative when function f(x,y) =Sin xy. both wrt x and y
Singularity
  • Singularity
Thinker: The derivative of sin(f(x,y)) w.r.t. x would be cos(f(x,y)) * df(x,y)/dx. In other words, take the derivative of cosine, then take the derivative of what's inside the cosine. Applying that formula, the d(sin(xy))/dx = cos(xy)y. Does that help?
anonymous
  • anonymous
i have differentiated w.r.t. x and got −y2Sinxy but to find 2nd order part. derv. i have differentiated this w.r.t. y and i have got −xySinxy−cosxy is this correct?
Singularity
  • Singularity
I'm not quite sure what '2nd order partial derivative' is (and I'm confused by your notation). But if you take the second derivative of sin(xy) w.r.t. x, you get -y^2*sin(xy) (negative y squared times the sine of x times y). The second deriv wrt y is -x^2*sin(xy). You can also take partial derivatives wrt x, then the partial derivative of that wrt to y.
anonymous
  • anonymous
yes i have got 1st derv(after diff wrt x, say eq 1) as wat you got, now i need to differentiate this eq. wrt y...i.e. eq 1 wrt 'y'
Singularity
  • Singularity
Then you'll need to use the product rule on cos(xy)y. For the product rule, u=cos(xy), and v=y. The product rule is (uv)'=u'v+uv' (the derivative of u times v equals the derivative of u times y plus u times the derivative of v. I think you have enough to complete the problem now.

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