Advertisements
Advertisements
प्रश्न
Find `dy/dx` if y = xx + 5x
उत्तर
y = xx + 5x
let u= xx and v = 5x
∴ y = u + v
∴ `dy/dx = (du)/dx + (dv)/dx` (i)
Now u = xx
Taking logarithm on both the sides
log u = log xx
⇒ log u = x.log x
Differentiate w.r.t.x.
`1/u (du)/dx = x. 1/x + log x.1`
= 1 + log x
`therefore dy/dx = u( 1 + log x )`
`therefore du/dx = xx( 1 + log x )` (ii)
Now, v = 5x
`therefore dv/dx = 5x.log 5` (iii)
Substituting (ii) and (iii) in eqn (i)
`dy/dx = x^x ( 1 + log x ) + 5^x log 5`
APPEARS IN
संबंधित प्रश्न
Differentiate the following function with respect to x: `(log x)^x+x^(logx)`
Differentiate the function with respect to x.
xsin x + (sin x)cos x
Differentiate the function with respect to x.
`(x cos x)^x + (x sin x)^(1/x)`
Find `dy/dx`for the function given in the question:
xy + yx = 1
Find `"dy"/"dx"` if y = xx + 5x
If x = esin3t, y = ecos3t, then show that `dy/dx = -(ylogx)/(xlogy)`.
If x = sin–1(et), y = `sqrt(1 - e^(2t)), "show that" sin x + dy/dx` = 0
If y = `log(x + sqrt(x^2 + a^2))^m`, show that `(x^2 + a^2)(d^2y)/(dx^2) + x "d"/"dx"` = 0.
Choose the correct option from the given alternatives :
If xy = yx, then `"dy"/"dx"` = ..........
If y = log [cos(x5)] then find `("d"y)/("d"x)`
If y = `(sin x)^sin x` , then `"dy"/"dx"` = ?
If `("f"(x))/(log (sec x)) "dx"` = log(log sec x) + c, then f(x) = ______.
Derivative of `log_6`x with respect 6x to is ______
`log [log(logx^5)]`
If y `= "e"^(3"x" + 7), "then the value" |("dy")/("dx")|_("x" = 0)` is ____________.
If y = `x^(x^2)`, then `dy/dx` is equal to ______.
The derivative of log x with respect to `1/x` is ______.
Find `dy/dx`, if y = (log x)x.
Find the derivative of `y = log x + 1/x` with respect to x.
