Question

Question #fbdf4

Answer 1

`d/dx5ln(5x) = 5/x`

Explanation:

Depends on what tools you have available.

Using the chain rule, along with the known derivative `d/dxlnx = 1/x`:

`d/dx5ln(5x) = 5d/dxln(5x)`

`=5(1/(5x))(d/dx5x)`

`=5(1/(5x))(5)`

`=5/x`

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Using implicit differentiation:

Let `y = 5ln(5x)`

`=> e^y = e^(5ln(5x)) = e^(ln((5x)^5))=(5x)^5=5^5x^5`

`=> d/dxe^y = d/dx5^5x^5`

`=> e^ydy/dx = 5^6x^4`

`=> dy/dx = (5^6x^4)/e^y`

`=(5^6x^4)/(5^5x^5)`

`=5/x`

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Using the definition of a derivative:

`lim_(h->0)(5ln(5(x+h))-5ln(5x))/h`

`=5lim_(h->0)(ln(5(x+h))-ln(5x))/h`

`=5lim_(h->0)ln((5(x+h))/(5x))/h`

`=5lim_(h->0)1/hln(1+h/x)`

`=5lim_(h->0)ln[(1+h/x)^(1/h)]`

`=5lim_(h->0)ln[(1+(1/x)/(1/h))^(1/h)]`

Substitute `u = 1/h`. Then `u->oo` as `h->0`. Recall `lim_(n->oo)(1+x/n)^n = e^x`.

`=5lim_(u->oo)ln[(1+(1/x)/u)^u]`

`=5ln(e^(1/x))`

`=5/xln(e)`

`=5/x`