# Thread: given f(x) and dy/dx with constant * f(x). K is?

1. ## given f(x) and dy/dx with constant * f(x). K is?

How do I go about tackling the problem below?
$\displaystyle {\rm If }y = 4^x {\rm then }\frac{{{\rm dy}}}{{{\rm dx}}} = k.4^x {\rm where k is: ?}$

Is started with trying first principles and came up with this
$\displaystyle \mathop {{\rm lim}}\limits_{{\rm h} \to {\rm 0}} \frac{{4^{a + h} - 4^a }}{h}$

I'm stuck

2. Originally Posted by Craka
How do I go about tackling the problem below?
$\displaystyle {\rm If }y = 4^x {\rm then }\frac{{{\rm dy}}}{{{\rm dx}}} = k.4^x {\rm where k is: ?}$

Is started with trying first principles and came up with this
$\displaystyle \mathop {{\rm lim}}\limits_{{\rm h} \to {\rm 0}} \frac{{4^{a + h} - 4^a }}{h}$

I'm stuck
$\displaystyle y=4^x=e^{\ln(4) x}$

So using the chain rule and the known derivative of $\displaystyle e^x$etc:

$\displaystyle \frac{dy}{dx}=\frac{d}{dx}\left[\ln(4) x \right]\ e^{\ln(4)x}$

which you should be able to complete

RonL

3. Originally Posted by Craka
How do I go about tackling the problem below?
$\displaystyle {\rm If }y = 4^x {\rm then }\frac{{{\rm dy}}}{{{\rm dx}}} = k.4^x {\rm where k is: ?}$

Is started with trying first principles and came up with this
$\displaystyle \mathop {{\rm lim}}\limits_{{\rm h} \to {\rm 0}} \frac{{4^{a + h} - 4^a }}{h}$

I'm stuck
if
y=a^x then
dy/dx=a^x(lna)
so if
y=4^x
dy/dx=4^x(ln4) but its given
dy/dx=k.4^x
comparing both you can get the value of k=ln4