Advertisements
Advertisements
рдкреНрд░рд╢реНрди
Find `bb(dy/dx)` for the given function:
xy = `e^((x - y))`
Advertisements
рдЙрддреНрддрд░
Given, xy = `e^((x - y))`
Taking logarithm of both the sides,
log (xy) = ` log e^((x - y))`
log x + log y = (x − y) loge e ....[тИ╡ log xy = log x + log y]
log x + log y = x − y ...[тИ╡ loge e = 1]
Differentiating both sides with respect to x,
`d/dx log x + d/dx log y = d/dx (x) - d/dx (y)`
`1/x + 1/y dy/dx = 1 - dy/dx `
`1/y dy/dx + dy/dx = 1 - 1/x`
`dy/dx ((1 + y)/y) = 1 - 1/x`
`((1 + y)/y) dy/dx = (x - 1)/x`
`therefore dy/dx = (y (x - 1))/(x (1 + y))`
APPEARS IN
рд╕рдВрдмрдВрдзрд┐рдд рдкреНрд░рд╢реНтАНрди
If `y=log[x+sqrt(x^2+a^2)]` show that `(x^2+a^2)(d^2y)/(dx^2)+xdy/dx=0`
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.
`(x + 1/x)^x + x^((1+1/x))`
Differentiate the function with respect to x.
`(sin x)^x + sin^(-1) sqrtx`
Find `bb(dy/dx)` for the given function:
(cos x)y = (cos y)x
Differentiate (x2 – 5x + 8) (x3 + 7x + 9) in three ways mentioned below:
- By using the product rule.
- By expanding the product to obtain a single polynomial.
- By logarithmic differentiation.
Do they all give the same answer?
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
xy = ex-y, then show that `"dy"/"dx" = ("log x")/("1 + log x")^2`
Find `"dy"/"dx"` if y = xx + 5x
If `log_5((x^4 + y^4)/(x^4 - y^4)) = 2, "show that""dy"/"dx" = (12x^3)/(13y^3)`.
If xy = ex–y, then show that `"dy"/"dx" = logx/(1 + logx)^2`.
`"If" y = sqrt(logx + sqrt(log x + sqrt(log x + ... ∞))), "then show that" dy/dx = (1)/(x(2y - 1).`
If y = `x^(x^(x^(.^(.^.∞))`, then show that `"dy"/"dx" = y^2/(x(1 - logy).`.
If ey = yx, then show that `"dy"/"dx" = (logy)^2/(log y - 1)`.
If x = `asqrt(secθ - tanθ), y = asqrt(secθ + tanθ), "then show that" "dy"/"dx" = -y/x`.
If x = esin3t, y = ecos3t, then show that `dy/dx = -(ylogx)/(xlogy)`.
Choose the correct option from the given alternatives :
If xy = yx, then `"dy"/"dx"` = ..........
If y = `25^(log_5sin_x) + 16^(log_4cos_x)` then `("d"y)/("d"x)` = ______.
If y = `log[sqrt((1 - cos((3x)/2))/(1 +cos((3x)/2)))]`, find `("d"y)/("d"x)`
If y = 5x. x5. xx. 55 , find `("d"y)/("d"x)`
The rate at which the metal cools in moving air is proportional to the difference of temperatures between the metal and air. If the air temperature is 290 K and the metal temperature drops from 370 K to 330 K in 1 O min, then the time required to drop the temperature upto 295 K.
lf y = `2^(x^(2^(x^(...∞))))`, then x(1 - y logx logy)`dy/dx` = ______
`d/dx(x^{sinx})` = ______
If y = `log ((1 - x^2)/(1 + x^2))`, then `"dy"/"dx"` is equal to ______.
`lim_("x" -> 0)(1 - "cos x")/"x"^2` is equal to ____________.
If y `= "e"^(3"x" + 7), "then the value" |("dy")/("dx")|_("x" = 0)` is ____________.
Given f(x) = `log((1 + x)/(1 - x))` and g(x) = `(3x + x^3)/(1 + 3x^2)`, then fog(x) equals
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 ______.
The derivative of x2x w.r.t. x is ______.
What is logarithmic differentiation?
For which type of function is logarithmic differentiation especially useful?
What is \[\frac{dy}{dx}\] for \[y=x^{\sin x}\], \[x>0\]?
