Advertisements
Advertisements
प्रश्न
Find `("d"y)/("d"x)`, if x = `sqrt(1 + "u"^2)`, y = log(1 +u2)
Advertisements
उत्तर
x = `sqrt(1 + "u"^2)`
Differentiating both sides w.r.t. u, we get
`("d"x)/"du" = "d"/"du"(sqrt(1 + "u"^2))`
= `1/(2sqrt(1 + "u"^2))*"d"/"du"(1 + "u"^2)`
= `1/(2sqrt(1 + "u"^2)) xx (0 + 2"u")`
= `"u"/sqrt(1 + "u"^2)`
y = log(1 + u2)
Differentiating both sides w.r.t. u, we get
`("d"y)/"du" = "d"/"du"[log(1 + "u"^2)]`
= `1/(1 + "u"^2)*"d"/"du"(1 + "u"^2)`
= `1/(1 + "u"^2) xx (0 + 2"u") = (2"u")/(1 + "u"^2)`
∴ `("d"y)/("d"x) = ((("d"y)/"du"))/((("d"x)/("du"))) = (((2"u")/(1 + "u"^2)))/(("u"/sqrt(1 + "u"^2))`
= `2/(1 + "u"^2) xx sqrt(1 + "u"^2)`
∴ `("d"y)/("d"x) = 2/sqrt(1 + "u"^2)`
APPEARS IN
संबंधित प्रश्न
Find `"dy"/"dx"`if, y = (2x + 5)x
Find `"dy"/"dx"`if, y = `root(3)(("3x" - 1)/(("2x + 3")(5 - "x")^2))`
Find `"dy"/"dx"`if, y = `10^("x"^"x") + 10^("x"^10) + 10^(10^"x")`
Fill in the blank.
If x = t log t and y = tt, then `"dy"/"dx"` = ____
Fill in the blank.
If y = y = [log (x)]2 then `("d"^2"y")/"dx"^2 =` _____.
State whether the following is True or False:
The derivative of `log_ax`, where a is constant is `1/(x.loga)`.
Find `"dy"/"dx"` if y = `sqrt(((3"x" - 4)^3)/(("x + 1")^4("x + 2")))`
Choose the correct alternative:
If y = (x )x + (10)x, then `("d"y)/("d"x)` = ?
If xy = 2x – y, then `("d"y)/("d"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 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)
`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 = x^(e^x)`
Find `dy/dx` if, `y = x^(e^x)`
Find `dy/dx "if", y = x^(e^x)`
Find `dy/(dx)` if, `x = e^(3t), y = e^sqrtt`.
