हिंदी

Find Dy/Dx If X^3 + Y^2 + Xy = 7 - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

Find `dy/dx if x^3 + y^2 + xy = 7`

योग
Advertisements

उत्तर

`x3 + y2 + xy = 7`
Differentiating both sides w.r.t.x.
`3x^2 + 2y dy/dx + x. dy/dx + y = 0`

`( 2y + x )dy/dx = -3x^2 - y`

`therefore dy/dx = [ - ( y + 3x^2 )]/[ 2y + x ]`

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
2018-2019 (March) Set 1

वीडियो ट्यूटोरियलVIEW ALL [3]

संबंधित प्रश्न

Is |sin x| differentiable? What about cos |x|?


Find `"dy"/"dx"` ; if x = sin3θ , y = cos3θ


Find `(dy)/(dx)` if `y = sin^-1(sqrt(1-x^2))`


If y = `sqrt(cosx + sqrt(cosx + sqrt(cosx + ... ∞)`, then show that `"dy"/"dx" = sinx/(1 - 2y)`.


If x = cos t, y = emt, show that `(1 - x^2)(d^2y)/(dx^2) - x"dy"/"dx" - m^2y` = 0.


If `sec^-1((7x^3 - 5y^3)/(7^3 + 5y^3)) = "m", "show"  (d^2y)/(dx^2)` = 0.


Find the nth derivative of the following:

`(1)/x`


Choose the correct option from the given alternatives :

If f(x) = `sin^-1((4^(x + 1/2))/(1 + 2^(4x)))`, which of the following is not the derivative of f(x)?


Choose the correct option from the given alternatives :

If y = sin (2sin–1 x), then dx = ........


If x = `e^(x/y)`, then show that `dy/dx = (x - y)/(xlogx)`


If log (x + y) = log (xy) + a then show that, `"dy"/"dx" = (- "y"^2)/"x"^2`.


State whether the following statement is True or False:

If `sqrt(x) + sqrt(y) = sqrt("a")`, then `("d"y)/("d"x) = 1/(2sqrt(x)) + 1/(2sqrt(y)) = 1/(2sqrt("a"))`


Let y = y(x) be a function of x satisfying `ysqrt(1 - x^2) = k - xsqrt(1 - y^2)` where k is a constant and `y(1/2) = -1/4`. Then `(dy)/(dx)` at x = `1/2`, is equal to ______.


If `tan ((x + y)/(x - y))` = k, then `dy/dx` is equal to ______.


If log (x+y) = log (xy) + a then show that, `dy/dx= (-y^2)/(x^2)`


If log(x + y) = log(xy) + a then show that, `dy/dx=(-y^2)/x^2`


If log(x + y) = log(xy) + a then show that, `dy/dx = (-y^2)/x^2`


Find `dy/(dx)  "if" , x = e^(3t), y = e^sqrtt`. 


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×