Question

Derive 5ln7x

Answer 1

Let's take a look, using the definition of derivative, how we can derive a function that is multiplied by a real number:
\[u(x)=kf(x)\]
by definition u'(x) must be:
\[u'(x)=\lim_{x \rightarrow a}\frac{ kf(x)-kf(a) }{ x-a }\]
taking "k" as common factor:
\[u'(x)=\lim_{x \rightarrow a}\frac{ k(f(x)-f(a)) }{ x-a }\]
And then splitting it as a product of limits:
\[u'(x)=\lim_{x \rightarrow a} \frac{ f(x)-f(a) }{ x-a }\lim_{x \rightarrow a}k\]
On the left side, we have the derivative of f, and on the right, the limit of a constant, wich is just the bare constant:
\[u'(x)=f'(x)k=kf'(x)\]
In conclusion:
\[u'(x)=kf'(x)\]
So that means, if I have a function, that consists of a number multiplying another function, that means that i just derive the function nad multiply it by the initial constant, so if we have:
\[f(x)=5Ln(7x)\]
then f'(x) must be:
\[f'(x)=5(\frac{ 1 }{ 7x })(7)\]