Question
Using property of logarithms, how do I prove derivative of ln(kx) is 1/x
First observe that ln(kx) = ln(k) + ln(x) then take derivatives. The ln(k) is simply a constant so it goes away. You could also derive it as
d/dx ln(kx) = 1/kx * k by the chain rule.
To see that the derivative of ln(x) is 1/x here's a brief proof.
If you have y=ln(x) then
e<sup>y</sup>=x Now find dx/dy to get
dx/dy = e<sup>y</sup> because
d/dy e<sup>y</sup> = e<sup>y</sup>
So 1/dx/dy = dy/dx = 1/e<sup>y</sup> = 1/x
This assumes that you know
d/du e<sup>u</sup> = e<sup>u</sup> which is easier to derive from the definition than d/dx ln(x).
what are you doing please give alghorthm problums
First observe that ln(kx) = ln(k) + ln(x) then take derivatives. The ln(k) is simply a constant so it goes away. You could also derive it as
d/dx ln(kx) = 1/kx * k by the chain rule.
To see that the derivative of ln(x) is 1/x here's a brief proof.
If you have y=ln(x) then
e<sup>y</sup>=x Now find dx/dy to get
dx/dy = e<sup>y</sup> because
d/dy e<sup>y</sup> = e<sup>y</sup>
So 1/dx/dy = dy/dx = 1/e<sup>y</sup> = 1/x
This assumes that you know
d/du e<sup>u</sup> = e<sup>u</sup> which is easier to derive from the definition than d/dx ln(x).
what are you doing please give alghorthm problums