Advertisements
Advertisements
प्रश्न
Find the second order derivative of the function.
x20
Advertisements
उत्तर
Let, y = x20
Differentiating both sides with respect to x,
`dy/dx = d/dx x^20`
= `20x^(20 - 1)`
= 20 x19
Differentiating both sides again with respect to x,
`(d^2 y)/dx^2 = 20 d/dx x^19`
= `20 xx 19x^(19 - 1)`
= 380 x18
APPEARS IN
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.
x . cos x
Find the second order derivative of the function.
x3 log x
Find the second order derivative of the function.
e6x cos 3x
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 y = 3 cos (log x) + 4 sin (log x), show that x2y2 + xy1 + y = 0.
If ey (x + 1) = 1, show that `(d^2y)/(dx^2) = (dy/dx)^2`.
If y = (tan–1 x)2, show that (x2 + 1)2 y2 + 2x (x2 + 1) y1 = 2
If x7 . y9 = (x + y)16 then show that `"dy"/"dx" = "y"/"x"`
Find `("d"^2"y")/"dx"^2`, if y = `"x"^5`
Find `("d"^2"y")/"dx"^2`, if y = `"x"^-7`
`sin xy + x/y` = x2 – y
tan–1(x2 + y2) = a
(x2 + y2)2 = xy
If y = tan–1x, find `("d"^2y)/("dx"^2)` in terms of y alone.
Derivative of cot x° with respect to x is ____________.
If y = `sqrt(ax + b)`, prove that `y((d^2y)/dx^2) + (dy/dx)^2` = 0.
If x = A cos 4t + B sin 4t, then `(d^2x)/(dt^2)` is equal to ______.
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))`
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 \[x\]?
For \[y=\mathrm{A}\sin x+\mathrm{B}\cos x\], what is \[\frac{d^2y}{dx^2}\]?
Which equation is satisfied by \[y=\mathrm{A}\sin x+\mathrm{B}\cos x\]?
If \[y=\sin^{-1}x\], what is \[\frac{dy}{dx}\]?
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 \[y=\sin^{-1}x\]?
What equation is obtained after multiplying through by \[\sqrt{1-x^2}\] in the derivation for \[y=\sin^{-1}x\]?
For \[y=\sin^{-1}x\], which relation uses \[y_{1}\] for the first derivative?
Differentiating \[(1-x^2)y_{1}^{2}=1\] gives which expression?
Which equation follows from \[(1-x^2)\cdot2y_{1}y_{2}+y_{1}^{2}(0-2x)=0\]?
