हिंदी

Find dydx, if x = 1+u2, y = log(1 +u2) - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

Find `("d"y)/("d"x)`, if x = `sqrt(1 + "u"^2)`, y = log(1 +u2)

योग
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)`

shaalaa.com
The Concept of Derivative - Derivatives of Logarithmic Functions
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 1.3: Differentiation - Q.4

संबंधित प्रश्न

Find `"dy"/"dx"`if, y = `"x"^("x"^"2x")`


Find `"dy"/"dx"`if, y = `"x"^("e"^"x")`


Find `"dy"/"dx"`if, y = `(log "x"^"x") + "x"^(log "x")`


Find `dy/dx`if, y = `(x)^x + (a^x)`.


If y = log `("e"^"x"/"x"^2)`, then `"dy"/"dx" = ?` 


Fill in the blank.

If x = t log t and y = tt, then `"dy"/"dx"` = ____


State whether the following is True or False:

If y = log x, then `"dy"/"dx" = 1/"x"`


State whether the following is True or False:

If y = e2, then `"dy"/"dx" = 2"e"`


Solve the following:

If y = [log(log(logx))]2, find `"dy"/"dx"`


If u = 5x and v = log x, then `("du")/("dv")` is ______


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 = 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)


Find `dy/dx  "if",y=x^(e^x) `


Find `dy/dx , if y^x = e^(x+y)`


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)`


Find `dy/(dx)  "if", y = x^(e^(x))` 


Find `dy/(dx)` if, `y = x^(e^x)`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×