# Find dydx, if y = xx + (7x – 1)x - Mathematics and Statistics

Sum

Find ("d"y)/("d"x), if y = xx + (7x – 1)x

#### Solution

y = xx + (7x – 1)x

Let u = xx and v = (7x – 1)

∴ y = u + v

Differentiating both sides w.r.t. x, we get

("d"y)/("d"x) = "du"/("d"x) + "dv"/("d"x)    ......(i)

Now, u = xx

Taking logarithm of both sides, we get

log u = log (xx)

∴ log u = x. log x

Differentiating both sides w.r.t. x, we get

"d"/("d"x)(log "u") = x*"d"/("d"x)(log x) + logx*"d"/("d"x)(x)

∴ 1/"u"*"du"/("d"x) = x*1/x + logx*1

∴ 1/"u"*"du"/("d"x) = 1 + log x

∴ "du"/("d"x) = u(1 + log x)

∴ "du"/("d"x) = xx(1 + log x)   ......(ii)

Also, v = (7x – 1)x

Taking logarithm of both sides, we get

log v = log(7x – 1)x

∴ log v = x.log(7x – 1)

Differentiating both sides w.r.t. x, we get

"d"/("d"x)(log "v") = x*"d"/("d"x)[log(7x - 1)] + log(7x - 1)*"d"/("d"x)(x)

∴ 1/"v"*"dv"/("d"x) = x*1/(7x - 1)*"d"/("d"x)(7x- 1) + log(7x - 1)*1

∴ 1/"v"*"dv"/("d"x) = x/(7x - 1)(7 - 0) + log(7x - 1)

∴ "dv"/("d"x) = "v"[(7x)/(7x - 1) + log(7x - 1)]

∴ "dv"/("d"x) = (7x - 1)^x[(7x)/(7x - 1) + log(7x - 1)]   .....(iii)

Substituting (ii) and (iii) in (i), we get

("d"y)/("d"x) = x^x(1 + logx) + (7x - 1)^x[log(7x - 1) + (7x)/(7x - 1)]

Concept: The Concept of Derivative - Derivatives of Logarithmic Functions
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Chapter 1.3: Differentiation - Q.5
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