Advertisements
Advertisements
प्रश्न
If y = elogx then `dy/dx` = ?
विकल्प
`(e^(logx))/x`
`1/x`
0
`1/2`
उत्तर
`(e^(log x))/x`
Explanation:
y = elogx
Differentiating both sides w.r.t.x, we get
`dy/dx = e^(logx).d/dx`(logx)
= `e^(logx).1/x`
= `(e^(logx))/x`
संबंधित प्रश्न
Find `"dy"/"dx"`if, y = `"x"^("x"^"2x")`
Find `"dy"/"dx"`if, y = `"e"^("x"^"x")`
Find `"dy"/"dx"`if, y = `(1 + 1/"x")^"x"`
Find `"dy"/"dx"`if, y = `root(3)(("3x" - 1)/(("2x + 3")(5 - "x")^2))`
Find `dy/dx`if, y = `(x)^x + (a^x)`.
Find `"dy"/"dx"`if, y = `10^("x"^"x") + 10^("x"^10) + 10^(10^"x")`
State whether the following is True or False:
If y = log x, then `"dy"/"dx" = 1/"x"`
The derivative of ax is ax log a.
Find `"dy"/"dx"` if y = `"x"^"x" + ("7x" - 1)^"x"`
If u = ex and v = loge x, then `("du")/("dv")` is ______
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 y = (log x)x + (x)logx
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)
If y = (log x)2 the `dy/dx` = ______.
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)`
Find `dy/dx` if, `y = x^(e^x)`
