Advertisements
Advertisements
प्रश्न
Find the second order derivative of the function.
x . cos x
Advertisements
उत्तर
Let, y = x cos x
Differentiating both sides with respect to x,
`dy/dx = x d/dx cos x + cos x d/dx (x)`
= x (−sin x) + cos x × 1
= −x sin x + cos x
Differentiating both sides again with respect to x,
`(d^2 y)/dx^2 = - [x d/dx sin x + sin x d/dx (x)] + d/dx cos x`
= −[x cos x + sin x × 1] + [−sin x]
= −x cos x − sin x − sin x
= −x cos x − 2 sin x
APPEARS IN
संबंधित प्रश्न
If y=2 cos(logx)+3 sin(logx), prove that `x^2(d^2y)/(dx2)+x dy/dx+y=0`
If x cos(a+y)= cosy then prove that `dy/dx=(cos^2(a+y)/sina)`
Hence show that `sina(d^2y)/(dx^2)+sin2(a+y)(dy)/dx=0`
Find the second order derivative of the function.
x2 + 3x + 2
Find the second order derivative of the function.
x20
Find the second order derivative of the function.
x3 log x
Find the second order derivative of the function.
ex sin 5x
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.
log (log x)
If y = cos–1 x, find `(d^2y)/dx^2` in terms of y alone.
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 = `"e"^"x"`
Find `("d"^2"y")/"dx"^2`, if y = log (x).
Find `("d"^2"y")/"dx"^2`, if y = 2at, x = at2
Find `("d"^2"y")/"dx"^2`, if y = `"x"^2 * "e"^"x"`
sec(x + y) = xy
If ax2 + 2hxy + by2 + 2gx + 2fy + c = 0, then show that `"dy"/"dx" * "dx"/"dy"` = 1
The derivative of cos–1(2x2 – 1) w.r.t. cos–1x is ______.
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 of the following is NOT a stated notation for the second derivative of \[f(x)\]?
Higher order derivatives are obtained by which process?
For \[y=\mathrm{A}\sin x+\mathrm{B}\cos x\], what is \[\frac{d^2y}{dx^2}\]?
After differentiating \[\sqrt{1-x^2}\cdot\frac{dy}{dx}=1\], which equation results?
What is \[\frac{d}{dx}\left(\sqrt{1-x^2}\right)\] in the differentiation for \[y=\sin^{-1}x\]?
