Find the derivative of [seventh root of x]^ln(x)?

Question

Find the derivative of [seventh root of x]^ln(x).

I have no idea where to start.

(x)^(1/7)^ln(x)? I get lost from here. Steps would be appreciated, as I'd like to learn how to actually do it. THank you so much.

Answer 1

2/7 x^(1/7 (-7 + Log[x])) Log[x]

Answer 2

One must use logarithmic differentiation:

Let y = {x^(1/7)}^ln(x)

A bit of simplification, first:

y = x^{(1/7)ln(x)}

Take the natural logarithm of both sides:

ln(y) = ln(x^{(1/7)ln(x)}

A bit more simplification before we differentiate:

ln(y) = (1/7)ln(x)ln(x)

ln(y) = (1/7){ln(x)}²

We must use the chain rule on both sides:

(1/y)(dy/dx) = (2/7x)ln(x)

dy/dx = (2/7x)ln(x)y

dy/dx = (2/7x)ln(x)[x^{(1/7)ln(x)}]

Answer 3

Note: This problem requires

taking ln of y twice.



Let y = x^(1/7)^ln(x)



Take ln of y:

ln(y) = ln(x^(1/7) ^ ln(x))

= (1/7)^ln(x) (ln(x))



Take ln of ln(y) because we cannot

directly take derivative of (1/7)^ln(x)

ln(ln(y)) = (ln(x)) (ln(1/7)) + ln(ln(x))



Take derivative of latter equation, using the chain rule:

(1/ln(y))(1/y)y' = (1/x) (ln(1/7)) + (1/ln(x))(1/x)



Solving for y' we have:

y' = (ln(y))(y) ((1/x) (ln(1/7)) + (1/ln(x))(1/x))

where y = x^(1/7)^ln(x) and ln(y) = ln(x^(1/7) ^ ln(x))



Simplified as much as possible we have:

y' = (1/7^ln(x)) (x^((1/7)^ln(x) - 1)) (ln(1/7)ln(x) +1)

Answer 4

Your question is derivative of x^(1/7)^(ln x).



Let y = x^(1/7)^(ln x)

Take log on both sides (to natural base),

log y = log (x) log (x^(1/7))

Differentiate w.r.t. x on both sides,

y'/y = log (x) [(1/x^(1/7)) × (1/7) × x^(-6/7)] + log (x^(1/7)) [1/x]

=> y'/y = log(x)/(7x) + log(x^(1/7))/(x)

=> y'/y = log(x)/(7x)+ log(x)/(7x)

=> y' = (2/7) y log(x)/x

=> y' = (2/7) x^(1/7)(log(x)-7) log(x)

Answer 5

Write f(x) = xlnx then the by-made from cos(f(x)) is what you're in seek of. d/dx cos(f(x)) = f'(x) * -sin(f(x), the popular thank you to distinguish cos. Use the product rule on f to discover its by-product: f'(x) = d/dx(x) * lnx + d/dx(lnx) * x = lnx + a million So the by-product is in simple terms -(lnx + a million)*sin(xlnx), no longer so frightening regardless of each and every thing.