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Question
Find `dy/dx`if, y = `(x)^x + (a^x)`.
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Solution
y = `(x)^x + (a^x)`
Let u = (x)x and v = (ax)
∴ y = u + v
Differentiating both sides w.r.t.x, we get
`dy/dx = (du)/dx + (dv)/dx` ....(i)
Now u = `(x)^x`
Taking logarithm of both sides, we get
log u = log `(x)^x`
∴ log u = x . log x
Differentiating both sides w.r.t.x, we get
`1/u (du)/dx = x * d/dx (log x) + log x * d/dx(x)`
`= x * 1/x + log x * (1)`
∴ `1/u (du)/dx = 1 + log x`
∴ `(du)/dx = u(1 + log x)`
∴ `(du)/dx = (x)^x` (1 + log x) ....(ii)
v = ax
Differentiating both sides w.r.t.x, we get
`(dv)/dx = a^x* log a` ....(iii)
Substituting (ii) and (iii) in (i), we get
`dy/dx = x^x(1 + log x) + a^x* log a`
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