हिंदी

Tan–1(x2 + y2) = a

Advertisements
Advertisements

प्रश्न

tan–1(x2 + y2) = a

योग
Advertisements

उत्तर

Given that: tan–1(x2 + y2) = a

⇒ x2 + y2 = tan a.

Differentiating both sides w.r.t. x.

`"d"/"dx"(x^2 + y^2) = "d"/"dx"(tan "a")`

⇒ `2x + 2y * "dy"/"dx"` = 0

⇒ `2y * "dy"/"dx"` = – 2x

⇒ `"dy"/"dx" = (-2x)/(2y) = (-x)/y`

Hence, `"dy"/"dx" = (-x)/y`.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 5: Continuity And Differentiability - Exercise [पृष्ठ १११]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 12
अध्याय 5 Continuity And Differentiability
Exercise | Q 56 | पृष्ठ १११

वीडियो ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्न

If x = a cos θ + b sin θ, y = a sin θ − b cos θ, show that `y^2 (d^2y)/(dx^2)-xdy/dx+y=0`


Find the second order derivative of the function.

ex sin 5x


Find the second order derivative of the function.

e6x cos 3x


If y = 3 cos (log x) + 4 sin (log x), show that x2y2 + xy1 + y = 0.


If y = Aemx + Benx, show that `(d^2y)/dx^2 - (m+ n) (dy)/dx + mny = 0`.


If y = 500e7x + 600e–7x, show that `(d^2y)/(dx^2)` = 49y.


If ey (x + 1) = 1, show that `(d^2y)/(dx^2) = (dy/dx)^2`.


If x7 . y9 = (x + y)16 then show that `"dy"/"dx" = "y"/"x"`


Find `("d"^2"y")/"dx"^2`, if y = `sqrt"x"`


Find `("d"^2"y")/"dx"^2`, if y = `"x"^5`


Find `("d"^2"y")/"dx"^2`, if y = `"x"^-7`


Find `("d"^2"y")/"dx"^2`, if y = `"e"^"x"`


sec(x + y) = xy


(x2 + y2)2 = xy


If ax2 + 2hxy + by2 + 2gx + 2fy + c = 0, then show that `"dy"/"dx" * "dx"/"dy"` = 1 


If x sin (a + y) + sin a cos (a + y) = 0, prove that `"dy"/"dx" = (sin^2("a" + y))/sin"a"`


If y = tan–1x, find `("d"^2y)/("dx"^2)` in terms of y alone.


The derivative of cos–1(2x2 – 1) w.r.t. cos–1x is ______.


If y = 5 cos x – 3 sin x, then `("d"^2"y")/("dx"^2)` is equal to:


Derivative of cot x° with respect to x is ____________.


If x2 + y2 + sin y = 4, then the value of `(d^2y)/(dx^2)` at the point (–2, 0) is ______.


Let for i = 1, 2, 3, pi(x) be a polynomial of degree 2 in x, p'i(x) and p''i(x) be the first and second order derivatives of pi(x) respectively. Let,

A(x) = `[(p_1(x), p_1^'(x), p_1^('')(x)),(p_2(x), p_2^'(x), p_2^('')(x)),(p_3(x), p_3^'(x), p_3^('')(x))]`

and B(x) = [A(x)]T A(x). Then determinant of B(x) ______


If x = A cos 4t + B sin 4t, then `(d^2x)/(dt^2)` is equal to ______.


If y = tan x + sec x then prove that `(d^2y)/(dx^2) = cosx/(1 - sinx)^2`.


Find `(d^2y)/dx^2 if, y = e^((2x + 1))`


Find `(d^2y)/dx^2` if, `y = e^((2x + 1))`


Find `(d^2y)/dx^2` if, y = `e^((2x + 1))`


Let \[y=f(x)\]. What is the first derivative of \[y\] with respect to \[x\]?


Which expression defines the second order derivative of \[y\] with respect to \[x\]?


For \[y=\mathrm{A}\sin x+\mathrm{B}\cos x\], what is \[\frac{d^2y}{dx^2}\]?


If \[y=\sin^{-1}x\], what is \[\frac{dy}{dx}\]?


Which equation is equivalent to \[\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}\] for \[y=\sin^{-1}x\]?


What is \[\frac{d}{dx}\left(\sqrt{1-x^2}\right)\] in the differentiation for \[y=\sin^{-1}x\]?


What equation is obtained after multiplying through by \[\sqrt{1-x^2}\] in the derivation for \[y=\sin^{-1}x\]?


Differentiating \[(1-x^2)y_{1}^{2}=1\] gives which expression?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×