Advertisements
Advertisements
प्रश्न
If ey ( x +1) = 1, then show that `(d^2 y)/(dx^2) = ((dy)/(dx))^2 .`
उत्तर
We have,
ey ( x +1) = 1
⇒ ey = `1/(x + 1)`
⇒ log `e^y = log (1/(x+1))`
⇒ y = - log (x + 1)
` ⇒ (dy)/(dx) = - 1/ (x + 1) and (d^2 y) /(dx^2) = 1/((x + 1)^2)`
` ⇒ (d^2 y)/(dx^2) = ((dy)/(dx))^2`
APPEARS IN
संबंधित प्रश्न
Differentiate the function with respect to x.
`sqrt(((x-1)(x-2))/((x-3)(x-4)(x-5)))`
Differentiate the function with respect to x.
`(log x)^(cos x)`
Differentiate the function with respect to x.
(x + 3)2 . (x + 4)3 . (x + 5)4
Find `dy/dx`for the function given in the question:
xy + yx = 1
If u, v and w are functions of x, then show that `d/dx(u.v.w) = (du)/dx v.w+u. (dv)/dx.w + u.v. (dw)/dx` in two ways-first by repeated application of product rule, second by logarithmic differentiation.
If x = a (cos t + t sin t) and y = a (sin t – t cos t), find `(d^2y)/dx^2`
Find `dy/dx` if y = xx + 5x
Find `"dy"/"dx"` , if `"y" = "x"^("e"^"x")`
Differentiate : log (1 + x2) w.r.t. cot-1 x.
Find `"dy"/"dx"` if y = xx + 5x
If y = (log x)x + xlog x, find `"dy"/"dx".`
If x = `asqrt(secθ - tanθ), y = asqrt(secθ + tanθ), "then show that" "dy"/"dx" = -y/x`.
If x = 2cos4(t + 3), y = 3sin4(t + 3), show that `"dy"/"dx" = -sqrt((3y)/(2x)`.
Differentiate 3x w.r.t. logx3.
Find the nth derivative of the following : log (ax + b)
If y = `25^(log_5sin_x) + 16^(log_4cos_x)` then `("d"y)/("d"x)` = ______.
Derivative of loge2 (logx) with respect to x is _______.
lf y = `2^(x^(2^(x^(...∞))))`, then x(1 - y logx logy)`dy/dx` = ______
If xy = ex-y, then `"dy"/"dx"` at x = 1 is ______.
`d/dx(x^{sinx})` = ______
`"d"/"dx" [(cos x)^(log x)]` = ______.
If `("f"(x))/(log (sec x)) "dx"` = log(log sec x) + c, then f(x) = ______.
If y = `("e"^"2x" sin x)/(x cos x), "then" "dy"/"dx" = ?`
`log (x + sqrt(x^2 + "a"))`
`log [log(logx^5)]`
If `"y" = "e"^(1/2log (1 + "tan"^2"x")), "then" "dy"/"dx"` is equal to ____________.
If y = `(1 + 1/x)^x` then `(2sqrt(y_2(2) + 1/8))/((log 3/2 - 1/3))` is equal to ______.
Evaluate:
`int log x dx`
