English

Differentiate the function with respect to x. (x + 3)2 . (x + 4)3 . (x + 5)4 - Mathematics

Advertisements
Advertisements

Question

Differentiate the function with respect to x.

(x + 3)2 . (x + 4)3 . (x + 5)4

Sum
Advertisements

Solution

Let, y = (x + 3)2 · (x + 4)3 · (x + 5)4

Taking logarithm of both sides,

log y = log [(x + 3)2 · (x + 4)3 · (x + 5)4]

= log (x + 3)2 + log (x + 4)3 + log (x + 5)4   ...[∵ log mn = log m + log n]

= 2 log (x + 3) + 3 log (x + 4) + 4 log (x + 5)  ...[∵ log mn = n log m]

Differentiating both sides with respect to x,

`1/y dy/dx = 2 d/dx log (x + 3) + 3 d/dx log (x + 4) + 4 d/dx log (x + 5)`

`1/y dy/dx = 2 * 1/(x + 3) d/dx (x + 3) + 3 xx 1/(x+ 4) d/dx (x + 4) + 4 xx 1/(x + 5) d/dx (x + 5)`

`1/y dy/dx = (2(1 + 0))/(x + 3) + (3(1 + 0))/("x" + 4) + (4(1 + 0))/(x + 5)`

`dy/dx = y [2/(x + 3) + 3/(x + 4) + 4/(x + 5)]`

`= y [(2 (x + 4) (x + 5) + 3 (x + 5) + 4 (x + 3) (x + 4))/((x + 3) (x + 4) (x + 5))]`

`= (x + 3)^2 (x + 4)^3 (x + 5)^4 xx [(2 (x^2 + 9x + 20) + 3(x^2 + 8x + 15) + 4 (x^2 + 7x + 12))/((x + 3) (x + 4) (x + 5))]`

`⇒ dy/dx= (x + 3) (x + 4)^2 (x + 5)^3 [9x^2 + 70x + 133]`

shaalaa.com
  Is there an error in this question or solution?
Chapter 5: Continuity and Differentiability - Exercise 5.5 [Page 178]

APPEARS IN

NCERT Mathematics Part 1 and 2 [English] Class 12
Chapter 5 Continuity and Differentiability
Exercise 5.5 | Q 5 | Page 178

RELATED QUESTIONS

Differentiate the following function with respect to x: `(log x)^x+x^(logx)`


Differentiate the function with respect to x.

(log x)cos x


Differentiate the function with respect to x.

xx − 2sin x


Differentiate the function with respect to x.

(log x)x + xlog x


Find `bb(dy/dx)` for the given function:

(cos x)y = (cos y)x


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.


Differentiate the function with respect to x:

xx + xa + ax + aa, for some fixed a > 0 and x > 0


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`.


Find `dy/dx` if y = x+ 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 `"x"^(5/3) . "y"^(2/3) = ("x + y")^(7/3)` , the show that `"dy"/"dx" = "y"/"x"`


If log (x + y) = log(xy) + p, where p is a constant, then prove that `"dy"/"dx" = (-y^2)/(x^2)`.


If `log_10((x^3 - y^3)/(x^3 + y^3))` = 2, show that `dy/dx = -(99x^2)/(101y^2)`.


If `log_5((x^4 + y^4)/(x^4 - y^4)) = 2, "show that""dy"/"dx" = (12x^3)/(13y^3)`.


`"If"  y = sqrt(logx + sqrt(log x + sqrt(log x + ... ∞))), "then show that"  dy/dx = (1)/(x(2y - 1).`


If ey = yx, then show that `"dy"/"dx" = (logy)^2/(log y - 1)`.


If x = esin3t, y = ecos3t, then show that `dy/dx = -(ylogx)/(xlogy)`.


If x = sin–1(et), y = `sqrt(1 - e^(2t)), "show that"  sin x + dy/dx` = 0


If x = `(2bt)/(1 + t^2), y = a((1 - t^2)/(1 + t^2)), "show that" "dx"/"dy" = -(b^2y)/(a^2x)`.


Find the second order derivatives of the following : x3.logx


Find the nth derivative of the following : log (2x + 3)


If y = A cos (log x) + B sin (log x), show that x2y2 + xy1 + y = 0.


If f(x) = logx (log x) then f'(e) is ______


Derivative of loge2 (logx) with respect to x is _______.


If y = `("e"^"2x" sin x)/(x cos x), "then" "dy"/"dx" = ?`


Derivative of `log_6`x with respect 6x to is ______


`2^(cos^(2_x)`


If xm . yn = (x + y)m+n, prove that `"dy"/"dx" = y/x`


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 ______.


If y = `x^(x^2)`, then `dy/dx` is equal to ______.


Derivative of log (sec θ + tan θ) with respect to sec θ at θ = `π/4` is ______.


If y = `9^(log_3x)`, find `dy/dx`.


Find the derivative of `y = log x + 1/x` with respect to x.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×