Advertisements
Advertisements
प्रश्न
Differentiate the function with respect to x.
cos x . cos 2x . cos 3x
Advertisements
उत्तर
Let, y = cos x · cos 2x · cos 3x .....(1)
Taking logarithm of both the sides,
log y = log (cos x · cos 2x · cos 3x)
log y = log cos x + log cos 2x + log cos 3x ....[∵ log m · n = log m + log n]
Differentiating both sides with respect to x,
`1/y dy/dx = d/dx log cos x + d/dx log cos 2x + d/dx log cos 3 x`
`1/y dy/dx = 1/(cos x) d/dx cos x + 1/(cos 2x) d/dx cos 2x + 1/(cos 3x) d/dx cos 3x`
`1/y dy/dx = 1/cos x (- sin x) + 1/(cos 2x) (- sin 2x) d/dx (2x) + 1/(cos 3x) (- sin 3x) d/dx (3x)`
`1/y dy/dx = - sin/cos x - (sin 2 x)/(cos 2 x) (2) - (sin 3 x)/(cos 3 x) (3)`
`1/y dy/dx` = − tan x − 2 tan 2x − 3 tan 3x
`1/y dy/dx` = −(tan x + 2 tan 2x + 3 tan 3x)
∴ `dy/dx` = −y(tan x + 2 tan 2x + 3 tan 3x)
Putting the value of y from equation (1)
`dy/dx` = cos x · cos 2x · cos 3x (tan x + 2 tan 2x + 3 tan 3x)
APPEARS IN
संबंधित प्रश्न
Differentiate the function with respect to x.
(log x)cos x
Find `bb(dy/dx)` for the given function:
xy + yx = 1
Find `bb(dy/dx)` for the given function:
xy = `e^((x - y))`
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 cos y = x cos (a + y), with cos a ≠ ± 1, prove that `dy/dx = cos^2(a+y)/(sin a)`.
If y = `e^(acos^(-1)x)`, −1 ≤ x ≤ 1, show that `(1- x^2) (d^2y)/(dx^2) -x dy/dx - a^2y = 0`.
If ey ( x +1) = 1, then show that `(d^2 y)/(dx^2) = ((dy)/(dx))^2 .`
Find `"dy"/"dx"` , if `"y" = "x"^("e"^"x")`
If `"x"^(5/3) . "y"^(2/3) = ("x + y")^(7/3)` , the show that `"dy"/"dx" = "y"/"x"`
If `(sin "x")^"y" = "x" + "y", "find" (d"y")/(d"x")`
If y = (log x)x + xlog x, find `"dy"/"dx".`
If x = a cos3t, y = a sin3t, show that `"dy"/"dx" = -(y/x)^(1/3)`.
If y = log (log 2x), show that xy2 + y1 (1 + xy1) = 0.
Find the nth derivative of the following : log (2x + 3)
If f(x) = logx (log x) then f'(e) is ______
If y = log [cos(x5)] then find `("d"y)/("d"x)`
If y = `log[sqrt((1 - cos((3x)/2))/(1 +cos((3x)/2)))]`, find `("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` = ______
`d/dx(x^{sinx})` = ______
If y = `("e"^"2x" sin x)/(x cos x), "then" "dy"/"dx" = ?`
Derivative of `log_6`x with respect 6x to is ______
`8^x/x^8`
`lim_("x" -> 0)(1 - "cos x")/"x"^2` is equal to ____________.
`lim_("x" -> -2) sqrt ("x"^2 + 5 - 3)/("x" + 2)` is equal to ____________.
If y `= "e"^(3"x" + 7), "then the value" |("dy")/("dx")|_("x" = 0)` is ____________.
If `f(x) = log [e^x ((3 - x)/(3 + x))^(1/3)]`, then `f^'(1)` is equal to
Given f(x) = `log((1 + x)/(1 - x))` and g(x) = `(3x + x^3)/(1 + 3x^2)`, then fog(x) equals
The derivative of x2x w.r.t. x is ______.
Find `dy/dx`, if y = (sin x)tan x – xlog x.
What is logarithmic differentiation?
What is the first step in the standard procedure for \[y=[u(x)]^{v(x)}\]?
What is \[\frac{dy}{dx}\] for \[y=x^{\sin x}\], \[x>0\]?
After taking logarithms in logarithmic differentiation, which rules are used to simplify products, quotients and powers before differentiation?
Which derivative correctly represents differentiating \[\ln y\] carefully?
What condition must be ensured for an expression inside logarithm?
