Whats the derivative of (ln x)^(1/5)?

Question

Please help me. I got 5 / x ((ln x)^(4/5)), but the answers section of my textbook says differently. Thanks.

Answer 1

Use the chain rule:

d/dx f(u) = d/du f(u) · du/dx

Now, let u = ln x, f(u) = u^(1/5).

Thus,
d/dx (ln x)^(1/5)

We substitute ln x = u
= d/dx u^(1/5)

Then, substitute u^(1/5) = f(u)
= d/dx f(u)

Then, use the chain rule formula,
= d/du f(u) · du/dx

Thus, f(u) = u^(1/5), u = ln x
= d/du u^(1/5) · d/dx ln x

Then,
= 1/5 u^(-4/5) · 1/x

We simplify,
= 1/(5u^(4/5) · x)

Substitute back u = ln x.
Therefore,
= 1/[5x(ln x)^(4/5)] = the answer

^_^

Answer 2

5 Ln X 1

Answer 3

The answer is: 1/ (5x*(ln x)^4/5)

d/dx ln x = 1/x

The power rule is: d x^n /dx = n*x^(n-1)

So, by the chain rule:

d/dx (ln x)^(1/5) = (1/5) (ln x)^((1/5)-1) d/dx ln x

= (1/5)*((ln x)^-4/5)*(1/x)

= 1/(5x*(ln x)^4/5)

Answer 4

Let y=(ln x)^1/5
Taking log on both sides,we get
ln y=(1/5)*ln(ln x)
Taking derivative on both sides w.r.t.x,we get
(1/y)(dy/dx)=(1/5)(1/ln x)(1/x)
dy/dx=(1/5)y(1/ln x)(1/x)
=(1/5)*(ln x)^(1/5)*(1/ln x)*(1/x)
=(1/5)*[ (ln x)^(-4/5)]*(1/x)
=1/{5x (ln x)^(4/5) }

Answer 5

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d/dy ln(y) = 1/y This is just one of those for which you need to remember

Answer 6

1/5(ln x) ^ -(4/5) (1/x)

Answer 7

d/dx ( (ln x)^(1/5) )

= 1/5 (ln x)^(1/5 - 1) * d/dx ( (ln x) )
= 1/5 (ln x)^(-4/5) * 1/x
=1/(5x (ln x)^(4/5) )