derivative of log functions?? (1 Viewer)

[RHINO7]

I can't get the working out correct for these questions please help-


i) x loge x

ii) loge(loge x)

iii)e^x lnx

iv) Find the equation of the tangent to the curve y= loge (x-1) at the point (1, loge 2)

 

[watatank]

I can't get the working out correct for these questions please help-


i) x.lnx

ii) ln(ln x)

iii)e^x lnx

iv) Find the equation of the tangent to the curve y= loge (x-1) at the point (1, loge 2)

i) use product rule.

d/dx(x.lnx) = x.(1/x) + (1)lnx = 1+lnx

ii) i *think* this is how youre sposed to do it. im sure there's a quicker way though. long ass version of chain rule

say y = ln(ln x)

also, let u = lnx

so y = lnu
u = lnx

dy/dx = (dy/du) * (du/dx)

= 1/u * 1/x

= 1/lnx *1/x

= 1/x.lnx

iii) product rule again
(d/dx)e^x lnx = e^x(1/x) + e^x.lnx = e^x( lnx + 1)


iv) youre sposed to do put the x co-ordinate of the point in the derivative of y= loge (x-1) and then you've got your gradient. from there you use point gradient formula to get the equation of the tangent. but i seem to get stuck when it comes to finding the derivative, so thats about all i can help on that one. sorry.

good luck with maths! :wave:

 

[Forbidden.]

...
ii) i *think* this is how youre sposed to do it. im sure there's a quicker way though. long ass version of chain rule

say y = ln(ln x)

also, let u = lnx

so y = lnu
u = lnx

dy/dx = (dy/du) * (du/dx)

= 1/u * 1/x

= 1/lnx *1/x

= 1/x.lnx

...

Maths. Extension 1 is shit, I avoid substitution for the sake of non-extension purity:

d/dx[ln(lnx)]

(Derivative of log_e x or ln x = f'(x) * 1/f(x) = f'(x)/f(x))
(f'(x) = 1/x and f(x) = ln x)

= 1/x * 1/lnx

= 1/xlnx

= 1/xlog_e x