Advertisements
Advertisements
प्रश्न
Differentiate the following w.r.t. x : (logx)x – (cos x)cotx
Advertisements
उत्तर
Let y = (log x)x – (cos x)cotx
Put u = (log x)x and v = (cos x)cotx
Then y = u – v
∴ `"dy"/"dx" = "du"/"dx" - "dv"/"dx"` ...(1)
Take u = (log x)x
∴ log u = log(log x)x = x.log(log x)
Differentiating both sides w.r.t. x, we get
`1/u."du"/"dx" = "d"/"dx"[x.log(logx)]`
= `x"d"/"dx"[log(logx)] + log(logx)."d"/"dx"(x)`
= `x xx 1/logx."d"/"dx"(logx) + log(logx) xx 1`
= `x xx 1/logx xx 1/x + log(logx)`
∴ `"du"/"dx" = u[1/logx + log(logx)]`
= `(logx)^x[1/logx + log(logx)]` ...(2)
Also v = (cos x)cotx
∴ log v = log(cos x)cotx = (cot x).(log cos x)
Differentiating both sides w.r.t. x, we get
`1/v."dv"/"dx" = "d"/dx"[(cotx).log(cosx)]`
= `(cotx)."d"/"dx"(log cosx) + (log cosx)."d"/"dx"(cotx)`
= `cotx xx 1/cosx."d"/"dx"(cosx) + (logcosx)(-"cosec"^2x)`
= `cotx xx 1/cosx xx (-sinx) - ("cosec"^2x)(logcosx)`
∴ `"dv"/"dx" = v[1/tanx xx (-tanx) - ("cosec"^2x)(logcosx)]`
= –(cos x)cotx[1 + (cosec2x)(log cos x)] ....(3)
From (1), (2) and (3), we get
∴ `"dy"/"dx" = (logx)^x[1/logx + log(logx)] + (cosx)^cotx[1 + ("cosec"^2x)(logcosx)]`.
APPEARS IN
संबंधित प्रश्न
Differentiate the following w.r.t. x:
(x3 – 2x – 1)5
Differentiate the following w.r.t.x: cos(x2 + a2)
Differentiate the following w.r.t.x: `5^(sin^3x + 3)`
Differentiate the following w.r.t.x: cos2[log(x2 + 7)]
Differentiate the following w.r.t.x: `sinsqrt(sinsqrt(x)`
Differentiate the following w.r.t.x: `log[sec (e^(x^2))]`
Differentiate the following w.r.t.x: [log {log(logx)}]2
Differentiate the following w.r.t.x:
sin2x2 – cos2x2
Differentiate the following w.r.t.x: (1 + 4x)5 (3 + x −x2)8
Differentiate the following w.r.t.x: `x/(sqrt(7 - 3x)`
Differentiate the following w.r.t.x:
`sqrt(cosx) + sqrt(cossqrt(x)`
Differentiate the following w.r.t.x: `(1 + sinx°)/(1 - sinx°)`
Differentiate the following w.r.t.x: log[tan3x.sin4x.(x2 + 7)7]
Differentiate the following w.r.t.x:
`log[a^(cosx)/((x^2 - 3)^3 logx)]`
Differentiate the following w.r.t. x:
`(x^2 + 2)^4/(sqrt(x^2 + 5)`
Differentiate the following w.r.t. x : cot–1(4x)
Differentiate the following w.r.t. x : `sin^-1(x^(3/2))`
Differentiate the following w.r.t. x : `sin^4[sin^-1(sqrt(x))]`
Differentiate the following w.r.t. x : `cot^-1[cot(e^(x^2))]`
Differentiate the following w.r.t. x :
`cos^-1[(3cos(e^x) + 2sin(e^x))/sqrt(13)]`
Differentiate the following w.r.t. x : `sin^-1((1 - x^2)/(1 + x^2))`
Differentiate the following w.r.t. x : cos–1(3x – 4x3)
Differentiate the following w.r.t. x :
`sin^(−1) ((1 − x^3)/(1 + x^3))`
Differentiate the following w.r.t.x:
`cot^-1((1 + 35x^2)/(2x))`
Differentiate the following w.r.t. x : `tan^-1((a + btanx)/(b - atanx))`
Differentiate the following w.r.t. x : `cot^-1((4 - x - 2x^2)/(3x + 2))`
Differentiate the following w.r.t. x: xe + xx + ex + ee.
Differentiate the following w.r.t. x : `[(tanx)^(tanx)]^(tanx) "at" x = pi/(4)`
Show that `"dy"/"dx" = y/x` in the following, where a and p are constants : `sec((x^5 + y^5)/(x^5 - y^5))` = a2
Show that `"dy"/"dx" = y/x` in the following, where a and p are constants : `cos^-1((7x^4 + 5y^4)/(7x^4 - 5y^4)) = tan^-1a`
Differentiate y = etanx w.r. to x
If y = sin−1 (2x), find `("d"y)/(""d"x)`
If f(x) is odd and differentiable, then f′(x) is
If y = `"e"^(1 + logx)` then find `("d"y)/("d"x)`
If y = `tan^-1[sqrt((1 + cos x)/(1 - cos x))]`, find `("d"y)/("d"x)`
Differentiate `sin^-1((2cosx + 3sinx)/sqrt(13))` w.r. to x
If f(x) = 3x - 2 and g(x) = x2, then (fog)(x) = ________.
Derivative of (tanx)4 is ______
Let f(x) = `(1 - tan x)/(4x - pi), x ne pi/4, x ∈ [0, pi/2]`. If f(x) is continuous in `[0, pi/2]`, then f`(pi/4)` is ______.
Solve `x + y (dy)/(dx) = sec(x^2 + y^2)`
Differentiate `tan^-1 (sqrt((3 - x)/(3 + x)))` w.r.t. x.
