(logx)^x + x^logx

Up board classes August 07, 2025

1.(logx)^x + x^logx

Solve:

मान लीजिए y = (logx)^x + x^logx
तथा मान लीजिए u = (logx)^x तथा v = (x^logx)

y = u + v
x के सापेक्ष अवकलन लेने पर,
dy/dx = du/dx + dv/dx ...........(i)

अब, u =( logx)^x
दोनों तरफ log लेने पर,

=⟩ logu = log (logx)^x
=⟩ logu = xlog (logx)

x के सापेक्ष अवकलन करने पर,
(1/u)du/dx = x d/dx(log x) + log(logx)d/dX(x)

=⟩ (1/u)du/dx = x/logx × 1/x + log(logx)
=⟩ du/dx = u[1/logx + log(logx)]
=⟩ du/dx = ( logx)^x[1/logx + log(logx)]

पुनः v = x^logx
दोनों तरफ का लघु गुणक लेने पर,
logv = logx ^logx =⟩ logv = (logx)(logx)
=⟩ logv = (logx)²

x के सापेक्ष अवकलन करने पर,
(1/v)dv/dx = 2logx d/dx (logx) = 2logx × 1/x
=⟩ dv/dx = v[2logx/x]
=⟩ dv/dx = x^logx[2logx/x]

du/dx तथा dv/dx का मान समी (i) में रखने पर,