Advertisements
Advertisements
प्रश्न
Find the derivatives of the following:
y = `x^(logx) + (logx)^x`
Advertisements
उत्तर
y = `x^(logx) + (logx)^x`
Let u = xlog x, v = (log x)x
log u = log xlog x
log u = (log x)(log x)
log u = (log x)2
`1/u * ("d"u)/("d"x) = 2 log x xx 1/x`
`("d"u)/("d"x) = 2u (logx)/x` ........(1)
v = (log x)x
log v = log (log x)x
log v = x log (log x)
`1/"v" * ("dv")/("d"x) = x xx 1/logx xx 1/x + log(log x) xx 1`
`("dv")/("d"x) = "v"[1/logx + log(logx)]` ........(2)
y = u + v
`("d"y)/("d"x) = ("d"u)/("d"x) + ("dv")/("d"x)`
`("d"y)/("d"x) = 2u logx/x + "v"[1/logx + log (log x)]`
By equation (1) and (2)
`("d"y)/("d"x) = 2 x^(logx) * logx/x + (logx)^x [1/logx + log(logx)]`
APPEARS IN
संबंधित प्रश्न
Find the derivatives of the following functions with respect to corresponding independent variables:
f(x) = x sin x
Find the derivatives of the following functions with respect to corresponding independent variables:
y = cos x – 2 tan x
Differentiate the following:
y = `"e"^sqrt(x)`
Differentiate the following:
f(t) = `root(3)(1 + tan "t")`
Differentiate the following:
y = cos (a3 + x3)
Differentiate the following:
y = `(x^2 + 1) root(3)(x^2 + 2)`
Differentiate the following:
f(x) = `x/sqrt(7 - 3x)`
Differentiate the following:
y = `sqrt(1 + 2tanx)`
Differentiate the following:
y = `"e"^(3x)/(1 + "e"^x`
Find the derivatives of the following:
y = `x^(cosx)`
Find the derivatives of the following:
`tan^-1 = ((6x)/(1 - 9x^2))`
Find the derivatives of the following:
`cos[2tan^-1 sqrt((1 - x)/(1 + x))]`
Find the derivatives of the following:
x = a (cos t + t sin t); y = a (sin t – t cos t)
Find the derivatives of the following:
`cos^-1 ((1 - x^2)/(1 + x^2))`
Find the derivatives of the following:
sin-1 (3x – 4x3)
Find the derivatives of the following:
If y = `(sin^-1 x)/sqrt(1 - x^2)`, show that (1 – x2)y2 – 3xy1 – y = 0
Choose the correct alternative:
`"d"/("d"x) (2/pi sin x^circ)` is
Choose the correct alternative:
If y = cos (sin x2), then `("d"y)/("d"x)` at x = `sqrt(pi/2)` is
Choose the correct alternative:
x = `(1 - "t"^2)/(1 + "t"^2)`, y = `(2"t")/(1 + "t"^2)` then `("d"y)/("d"x)` is
