Advertisements
Advertisements
प्रश्न
Find `"dy"/"dx"`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)]`
`= 1/3 log ("3x" - 1) - [1/3 log ("2x" + 3) + 2/3 log (5 - "x")]`
Differentiating both sides w.r.t. x, we get
`1/"y" "dy"/"dx" = 1/3 * "d"/"dx" [log ("3x" - 1)] - 1/3 * "d"/"dx" [log (2"x" + 3)] - 2/3 * "d"/"dx" [log (5 - "x")]`
`= 1/3 * 1/("3x" - 1)*"d"/"dx" ("3x" - 1) - 1/3 * 1/("2x + 3") * "d"/"dx" ("2x" + 3) - 2/3 * 1/("5 - x") * "d"/"dx" (5 - "x")`
`= 1/(3(3"x" - 1)) xx 3 - 1/(3(2"x" + 3)) xx 2 - 2/(3(5 - "x")) xx -1`
∴ `1/"y" "dy"/"dx" = 1/("3x" - 1) - 2/(3("2x" + 3)) + 2/(3(5 - "x"))`
∴ `"dy"/"dx" = "y"/3 [3/("3x" - 1) - 2/("2x" + 3) + 2/(5 - "x")]`
∴ `"dy"/"dx" = 1/3 * root(3)(("3x" - 1)/(("2x + 3")(5 - "x")^2)) [3/("3x" - 1) - 2/("2x" + 3) + 2/(5 - "x")]`
APPEARS IN
संबंधित प्रश्न
Find `"dy"/"dx"`if, y = (2x + 5)x
Find `"dy"/"dx"`if, y = `(log "x"^"x") + "x"^(log "x")`
Find `"dy"/"dx"`if, y = `10^("x"^"x") + 10^("x"^10) + 10^(10^"x")`
If y = x log x, then `(d^2y)/dx^2`= ______.
Solve the following:
If y = [log(log(logx))]2, find `"dy"/"dx"`
Find `"dy"/"dx"` if y = `"x"^"x" + ("7x" - 1)^"x"`
Find `("d"y)/("d"x)`, if y = [log(log(logx))]2
Find `("d"y)/("d"x)`, if xy = log(xy)
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 = xx + (7x – 1)x
Find `("d"y)/("d"x)`, if y = `x^(x^x)`
Find `("d"y)/("d"x)`, if y = `root(3)(((3x - 1))/((2x + 3)(5 - x)^2)`
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)
Solve the following differential equations:
x2ydx – (x3 – y3)dy = 0
`int 1/(4x^2 - 1) dx` = ______.
If y = x . log x then `dy/dx` = ______.
If y = (log x)2 the `dy/dx` = ______.
FInd `dy/dx` if,`x=e^(3t), y=e^sqrtt`
Find `dy/dx "if", y = x^(e^x)`
Find `dy/dx` if, y = `x^(e^x)`
Find `dy/dx, "if" y=sqrt((2x+3)^5/((3x-1)^3(5x-2)))`
Find `dy/dx,"if" y=x^x+(logx)^x`
Find `dy/dx` if, `y = x^(e^x)`
