How do I find the derivative of log_base3 (1+ xln3)?

Question

If you can explain little bit, that would be even better
thank you

Answer 1

Alright, here's the derivatives you need to know:
d/dx(log_b (x)) = 1/(ln(b)x) (b is the base of the log)
d/dx(ax) = a
and you need to know the chain rule: d(f(g(x)))/dx = f'(g(x))*g'(x)

So by chain rule:
d/dx(log_3 (1+ xln3))=
1/(ln(3)*(1+xln3)) * d/dx(1+x*ln3)=
1/(ln(3)*(1+xln3)) * (0 + ln3)=
ln(3)/(ln(3)*(1+xln(3)))=
1/(1+xln(3)

Answer 2

The derivitive of one is zero. With xln(3), you must use the product rule here. The first times the derivative of the second, plus the second times the derivative of the first.

y`=x*0+1*ln(3)

y`=0+ln(3)

y`=ln(3)

Answer 3

First, remember that the derivative of ln(x) = 1/x, and that log base 3 (z) = ln(z)/ln(3)

Chain rule all the way:

d/dx (log base 3 (1 + xln(3)))

= d/dx (ln(1+xln(3))/ln(3))

(1/(ln(3) * (1+xln(3)))) * ln(3) = 1/(1+xln(3))

Answer 4

You may try another brand

let y = Log(at 3) {1 + x Log(at e) (3) }

take both sides to the base " 3" like we take exponential

(3)^y = {1 + x Log(at e) (3) } ... (1)

differentiate wrt x
(3)^y Log (at e) (3) . dy/dx = 0 + 1* Log(at e) (3)

(3)^y . dy/dx = 1

dy/dx = 1/ (3)^y = 1 / {1 + x Log(at e) (3) } from (1)

Answer - do it differently