Advertisements
Advertisements
प्रश्न
Differentiate the following w.r.t. x :
(sin x)tanx + (cos x)cotx
Advertisements
उत्तर
Let y = (sin x)tanx + (cos x)cotx
Put u = (sin x)tanx and v = (cos x)cotx
Then y = u + v
∴ `"dy"/"dx" = "du"/"dx" + "dv"/"dx"` ...(1)
Take u = (sin x)tanx
∴ log u = log(sin x)tanx = (tan x).(log sinx)
Differentiating both sides w.r.t. x, we get
`1/u."du"/"dx" = "d"/"dx"[(tan x)(log sin x)]`
= `(tan x)."d"/"dx"(log sin x) + (log sinx)."d"/"dx"(tanx)`
= `(tanx)/(sin x)."d"/"dx"(sin x) + (log sinx)(sec^2x)`
= `((sinx)/(cosx))/(sinx).cosx + (sec^2x)(log sinx)`
= 1 + (sec2x)(log sinx)
∴ `"du"/"dx" = y[1 + (sec^2x)(log sinx)]`
= (sin x)tanx[1 + (sec2x)(log sinx)] ...(2)
Also, v = (cos x)cotx
∴ log v = log(cos x)cotx = (cot x).(log cosx)
Differentiating both sides w.r.t. x, we get
`1/v."dv"/"dx" = "d"/"dx"[(cot x).(log cos x)]`
= `(cot x)."d"/"dx"(log cos x) + (log cos x)."d"/"dx"(cotx)`
= `cot x xx 1/cosx."d"/"dx"(cosx) + (log cosx).(-"cosec"^2x)`
= `cotx xx 1/cosx xx (-sin x) - ("cosec"^2x)(log cosx)`
∴ `"dv"/"dx" = v[1/tanx xx (-tanx) - ("cosec"^2x)(log cosx)]`
= –(cos x)cotx [1 + (cosec2x)(log cosx)] ...(3)
From (1), (2) and (3), we get
`"dy"/"dx" = (sin x)^(tanx)[1 + (sec^2x)(log sin x)] - (cos x)^(cotx)[1 + ("cosec"^2x)(log cosx)]`.
APPEARS IN
संबंधित प्रश्न
Differentiate the following w.r.t.x: `"cosec"(sqrt(cos x))`
Differentiate the following w.r.t.x:
tan[cos(sinx)]
Differentiate the following w.r.t.x: `log[sec (e^(x^2))]`
Differentiate the following w.r.t.x:
(x2 + 4x + 1)3 + (x3− 5x − 2)4
Differentiate the following w.r.t.x:
`(x^3 - 5)^5/(x^3 + 3)^3`
Differentiate the following w.r.t.x: `log(sqrt((1 - sinx)/(1 + sinx)))`
Differentiate the following w.r.t.x: `log[4^(2x)((x^2 + 5)/(sqrt(2x^3 - 4)))^(3/2)]`
Differentiate the following w.r.t. x : tan–1(log x)
Differentiate the following w.r.t. x : cosec–1 (e–x)
Differentiate the following w.r.t. x : cot–1(x3)
Differentiate the following w.r.t. x : `tan^-1(sqrt(x))`
Differentiate the following w.r.t. x : `sin^-1(x^(3/2))`
Differentiate the following w.r.t. x : `tan^-1[(1 - tan(x/2))/(1 + tan(x/2))]`
Differentiate the following w.r.t.x:
tan–1 (cosec x + cot x)
Differentiate the following w.r.t. x : `sin^-1((cossqrt(x) + sinsqrt(x))/sqrt(2))`
Differentiate the following w.r.t. x :
`cos^-1((1 - x^2)/(1 + x^2))`
Differentiate the following w.r.t. x : `tan^-1((2x)/(1 - x^2))`
Differentiate the following w.r.t. x : `sin^-1((1 - x^2)/(1 + x^2))`
Differentiate the following w.r.t. x : `cos^-1((e^x - e^(-x))/(e^x + e^(-x)))`
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 :
`tan^-1((5 -x)/(6x^2 - 5x - 3))`
Differentiate the following w.r.t. x :
`(x + 1)^2/((x + 2)^3(x + 3)^4`
Differentiate the following w.r.t. x : `root(3)((4x - 1)/((2x + 3)(5 - 2x)^2)`
Differentiate the following w.r.t. x : `(x^2 + 3)^(3/2).sin^3 2x.2^(x^2)`
Differentiate the following w.r.t. x : `(x^5.tan^3 4x)/(sin^2 3x)`
Differentiate the following w.r.t. x: (sin xx)
Differentiate the following w.r.t. x : `x^(e^x) + (logx)^(sinx)`
Differentiate the following w.r.t. x :
etanx + (logx)tanx
If f(x) is odd and differentiable, then f′(x) is
If y = `sqrt(cos x + sqrt(cos x + sqrt(cos x + ...... ∞)`, show that `("d"y)/("d"x) = (sin x)/(1 - 2y)`
y = {x(x - 3)}2 increases for all values of x lying in the interval.
A particle moves so that x = 2 + 27t - t3. The direction of motion reverses after moving a distance of ______ units.
If y = cosec x0, then `"dy"/"dx"` = ______.
Solve `x + y (dy)/(dx) = sec(x^2 + y^2)`
If y = log (sec x + tan x), find `dy/dx`.
`lim_(x → 0) (sqrt(1 + x + x^2) − 1)/x` = ______.
