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प्रश्न
Differentiate the following w.r.t. x:
`x^(x^x) + e^(x^x)`
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उत्तर
Let y = `x^(x^x) + e^(x^x)`
Put u = `x^(x^x) and v = e^(x^(x)`
Then y = u + v
∴ `"dy"/"dx" = "du"/"dx" + "dv"/"dx"` ...(1)
Take u = `x^(x^(x)`
∴ log u = `logx^(x^(x)) = x^x*logx`
Differentiating both sides w.r.t. x, we get
`1/u*"du"/"dx" = "d"/"dx"(x^x*logx)`
= `x^x*"d"/"dx"(logx) + (logx)*"d"/"dx"(x^x)`
= `x^x xx 1/x + (logx)*"d"/"dx"(x^x)` ...(2)
To find `"d"/"dx"(x^x)`
Let ω = xx
Then log ω = xlogx
Differentiating both sides w.r.t. x, we get
`1/omega*"dω"/"dx" = "d"/"dx"(xlogx)`
= `x*"d"/"dx"(logx) + (logx)*"d"/"dx"(x)`
= `x xx (1)/x + (logx) xx 1`
∴ `"dω"/"dx" = omega(1 + logx)`
∴ `"d"/"dx"(x^x) = x^x(1 + logx)` ...(3)
∴ from (2),
`1/u*"du"/"dx" = x^x xx (1)/x + (logx)*x^x(1 + logx)`
∴ `"du"/"dx" = y[x^x xx 1/x + (logx)*x^x(1 + logx)]`
= `x^(x^x)*x^x[1/x + (logx)*(1 + logx)]`
= `x^(x^x)*x^x*logx[1 + logx + 1/(xlogx)]` ...(4)
Also, v = `e^(x^(x)`
∴ `"dv"/"dx" = "d"/"dx"(e^(x^x))`
= `e^(x^(x))*"d"/"dx"(e^(x^x))`
= `e^(x^(x))*x^x(1 + logx)` ...(5) [By (3)]
From (1), (4) and (5), we get
`"dy"/"dx" = x^(x^x)*x^x*logx[1 + logx + 1/(xlogx)] + e^(x^x)*x^x(1 + logx)`
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