### 6 Answers

- MyRankLv 62 weeks ago
y = log (logx)

dy/dx = 1/ logx d/dx (logx)= 1/ logx x 1/x= 1/ xlogx.

Source(s): https://myrank.co.in/ - hfshawLv 72 weeks ago
You have:

y = ln(ln x)

There are many ways to find dy/dx (e.g., chain rule, implicit differentiation...). Let's use implicit differentiation.

Exponentiate both sides:

exp(y) = ln(x)

Differentiate both sides with respect to x:

exp(y)*dy/dx = 1/x

solve for dy/dx:

dy/dx = exp(-y)/x

Back substitute for exp(y) [if exp(y) = ln(x), exp(-y) = 1/ln(x)]:

dy/dx = 1/(x*ln(x))

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- Jim MoorLv 72 weeks ago
Find the derivative of the following via implicit differentiation:

d/dx(y) = d/dx(ln(ln(x)))

Using the chain rule, d/dx(y) = ( dy(u))/( du) ( du)/( dx), where u = x and d/( du)(y(u)) = y'(u):

d/dx(x) y'(x) = d/dx(ln(ln(x)))

The derivative of x is 1:

1 y'(x) = d/dx(ln(ln(x)))

INTERMEDIATE STEPS:

Possible derivation:

d/dx(x)

Use the power rule, d/dx(x^n) = n x^(n - 1), where n = 1.

d/dx(x) = d/dx(x^1) = x^0:

Answer: |

| = 1 |

Using the chain rule, d/dx(ln(ln(x))) = ( dln(u))/( du) ( du)/( dx), where u = ln(x) and d/( du)(ln(u)) = 1/u:

y'(x) = (d/dx(ln(x)))/ln(x)

Using the chain rule, d/dx(ln(x)) = ( dln(u))/( du) ( du)/( dx), where u = x and d/( du)(ln(u)) = 1/u:

y'(x) = ((d/dx(x))/x)/ln(x)

Simplify the expression:

y'(x) = (d/dx(x))/(x ln(x))

The derivative of x is 1:

Answer: y'(x) = 1 1/(x ln(x))