Advertisements
Advertisements
प्रश्न
Find `("d"y)/("d"x)`, if y = `root(3)(((3x - 1))/((2x + 3)(5 - x)^2)`
Advertisements
उत्तर
y = `root(3)(((3x - 1))/((2x + 3)(5 - x)^2)`
= `(3x - 1)^(1/3)/((2x + 3)^(1/3)*(5 - x)^(2/3)`
Taking logarithm of both sides, we get
log y = `log[(3x - 1)^(1/3)/((2x - 3)^(1/3)*(5 - x)^(2/3))]`
= `log(3x - 1)^(1/3) - [log(2x + 3)^(1/3) + log(5 - x)^(2/3)]`
= `1/3log(3x - 1) - [1/3 log(2x + 3) + 2/3log(5 - x)]`
Differentiating both sides w.r.t. x, we get
`"d"/("d"x)(log y) = 1/3*"d"/("d"x)[log(3x - 1)] - 1/3*"d"/("d"x)[log(2x + 3)] - 2/3*"d"/("d"x)[log(5 - x)]`
∴ `1/y*("d"y)/("d"x) = 1/3*1/((3x - 1))*"d"/("d"x)(3x - 1) - 1/3*1/((2x + 3))*"d"/("d"x)(2x + 3) - 2/3*1/((5 - x))*"d"/("d"x)(5 - x)`
∴ `1/y*("d"y)/("d"x) = 1/(3(3x - 1)) xx (3 - 0) - 1/(3(2x + 3)) xx (2 + 0) - 2/(3(5 - x)) xx (0 - 1)`
∴ `1/y*("d"y)/("d"x) = 1/(3x - 1)- 2/(3(2x + 3)) + 2/(3(5 - x))`
∴ `("d"y)/("d"x) = y/3[3/(3x - 1) - 2/(2x + 3) + 2/(5 - x)]`
∴ `("d"y)/("d"x) = 1/3 root(3)((3x - 1)/((2x + 3)*(5 - x)^2)) [3/(3x - 1) - 2/(2x + 3) + 2/(5 - x)]`
संबंधित प्रश्न
Find `"dy"/"dx"`if, y = `"x"^("e"^"x")`
Find `"dy"/"dx"`if, y = `root(3)(("3x" - 1)/(("2x + 3")(5 - "x")^2))`
Find `dy/dx`if, y = `(x)^x + (a^x)`.
If y = elogx then `dy/dx` = ?
If y = log `("e"^"x"/"x"^2)`, then `"dy"/"dx" = ?`
Fill in the Blank
If 0 = log(xy) + a, then `"dy"/"dx" = (-"y")/square`
The derivative of ax is ax log a.
Find `"dy"/"dx"` if y = `sqrt(((3"x" - 4)^3)/(("x + 1")^4("x + 2")))`
Differentiate log (1 + x2) with respect to ax.
Find `("d"y)/("d"x)`, if x = `sqrt(1 + "u"^2)`, y = log(1 +u2)
If x = t.logt, y = tt, then show that `("d"y)/("d"x)` = tt
Find `("d"y)/("d"x)`, if y = (log x)x + (x)logx
If xa .yb = `(x + y)^((a + b))`, then show that `("d"y)/("d"x) = y/x`
Find `("d"y)/("d"x)`, if y = x(x) + 20(x)
Solution: Let y = x(x) + 20(x)
Let u = `x^square` and v = `square^x`
∴ y = u + v
Diff. w.r.to x, we get
`("d"y)/("d"x) = square/("d"x) + "dv"/square` .....(i)
Now, u = xx
Taking log on both sides, we get
log u = x × log x
Diff. w.r.to x,
`1/"u"*"du"/("d"x) = x xx 1/square + log x xx square`
∴ `"du"/("d"x)` = u(1 + log x)
∴ `"du"/("d"x) = x^x (1 + square)` .....(ii)
Now, v = 20x
Diff.w.r.to x, we get
`"dv"/("d"x") = 20^square*log(20)` .....(iii)
Substituting equations (ii) and (iii) in equation (i), we get
`("d"y)/("d"x)` = xx(1 + log x) + 20x.log(20)
`int 1/(4x^2 - 1) dx` = ______.
If y = (log x)2 the `dy/dx` = ______.
Find`dy/dx if, y = x^(e^x)`
FInd `dy/dx` if,`x=e^(3t), y=e^sqrtt`
Find `dy/dx "if", y = x^(e^x)`
Find `dy/dx,"if" y=x^x+(logx)^x`
Find `dy/dx` if, y = `x^(e^x)`
Find `dy / dx` if, `y = x^(e^x)`
Find `dy/dx` if, y = `x^(e^x)`
Find `dy/dx` if, `y = x^(e^x)`
Find `dy/dx` if, `y = x^(e^x)`
Find `dy/(dx)` if, `y = x^(e^x)`
