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Use quantifiers to convert the given open sentence defined on N into a true statement.
3x – 4 < 9
Concept: undefined >> undefined
Use quantifiers to convert the given open sentence defined on N into a true statement.
Y + 4 > 6
Concept: undefined >> undefined
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Choose the correct alternative:
If y = (x )x + (10)x, then `("d"y)/("d"x)` = ?
Concept: undefined >> undefined
If xy = 2x – y, then `("d"y)/("d"x)` = ______
Concept: undefined >> undefined
If y = `"a"^((1 + log"x"))`, then `("d"y)/("d"x)` is ______
Concept: undefined >> undefined
If u = 5x and v = log x, then `("du")/("dv")` is ______
Concept: undefined >> undefined
If u = ex and v = loge x, then `("du")/("dv")` is ______
Concept: undefined >> undefined
State whether the following statement is True or False:
If y = log(log x), then `("d"y)/("d"x)` = logx
Concept: undefined >> undefined
State whether the following statement is True or False:
If y = 4x, then `("d"y)/("d"x)` = 4x
Concept: undefined >> undefined
Find `("d"y)/("d"x)`, if y = [log(log(logx))]2
Concept: undefined >> undefined
Find `(dy)/(dx)`, if xy = yx
Concept: undefined >> undefined
Find `("d"y)/("d"x)`, if xy = log(xy)
Concept: undefined >> undefined
Find `("d"y)/("d"x)`, if x = `sqrt(1 + "u"^2)`, y = log(1 +u2)
Concept: undefined >> undefined
If x = t.logt, y = tt, then show that `("d"y)/("d"x)` = tt
Concept: undefined >> undefined
Find `("d"y)/("d"x)`, if y = (log x)x + (x)logx
Concept: undefined >> undefined
Find `("d"y)/("d"x)`, if y = xx + (7x – 1)x
Concept: undefined >> undefined
Find `("d"y)/("d"x)`, if y = `x^(x^x)`
Concept: undefined >> undefined
Find `("d"y)/("d"x)`, if y = `root(3)(((3x - 1))/((2x + 3)(5 - x)^2)`
Concept: undefined >> undefined
If xa .yb = `(x + y)^((a + b))`, then show that `("d"y)/("d"x) = y/x`
Concept: undefined >> undefined
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)
Concept: undefined >> undefined
