Advertisements
Advertisements
प्रश्न
If y = `sqrt(ax + b)`, prove that `y((d^2y)/dx^2) + (dy/dx)^2` = 0.
Advertisements
उत्तर
Given, y = `sqrt(ax + b)`
Then, `dy/dx = a/(2sqrt(ax + b)`
`\implies dy/dx = a/(2y)`
`\implies y dy/dx = a/2`
Again, differentiating with respect to x, we get
`\implies y(d^2y)/(dx^2) + dy/dx xx dy/dx` = 0
`\implies y(d^2y)/(dx^2) + (dy/dx)^2` = 0
Hence Proved.
संबंधित प्रश्न
Find the second order derivative of the function.
e6x cos 3x
Find the second order derivative of the function.
tan–1 x
Find the second order derivative of the function.
sin (log x)
If y = 5 cos x – 3 sin x, prove that `(d^2y)/(dx^2) + y = 0`.
If y = cos–1 x, find `(d^2y)/dx^2` in terms of y alone.
If y = Aemx + Benx, show that `(d^2y)/dx^2 - (m+ n) (dy)/dx + mny = 0`.
If `x^3y^5 = (x + y)^8` , then show that `(dy)/(dx) = y/x`
Find `("d"^2"y")/"dx"^2`, if y = `"e"^((2"x" + 1))`.
Find `("d"^2"y")/"dx"^2`, if y = log (x).
Find `("d"^2"y")/"dx"^2`, if y = `"x"^2 * "e"^"x"`
`sin xy + x/y` = x2 – y
tan–1(x2 + y2) = a
(x2 + y2)2 = xy
If ax2 + 2hxy + by2 + 2gx + 2fy + c = 0, then show that `"dy"/"dx" * "dx"/"dy"` = 1
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 y = tan x + sec x then prove that `(d^2y)/(dx^2) = cosx/(1 - sinx)^2`.
If x = a cos t and y = b sin t, then find `(d^2y)/(dx^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))`
Which expression defines the second order derivative of \[y\] with respect to \[x\]?
Higher order derivatives are obtained by which process?
Which equation is equivalent to \[\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}\] for \[y=\sin^{-1}x\]?
Differentiating \[(1-x^2)y_{1}^{2}=1\] gives which expression?
Which equation follows from \[(1-x^2)\cdot2y_{1}y_{2}+y_{1}^{2}(0-2x)=0\]?
