Hi,
The following function is given:
\(
y(x) = \frac{1}{(b-ax)}
\)
I want to find the derivative of this one but I am stuck between two methods:
\(
y(x) = \frac{1}{(b-ax)}
y(x) = (b-ax)^{-1}
y(x) = b^{-1}-a^{-1}x^{-1}
y'(x) = 0 - a^{-1}*-x^{-2}
y'(x) = \frac{1}{a}*\frac{1}{x^{2}}
y'(x) = \frac{1}{ax^{2}}
\)
On the otherhand using the chain rule:
\(
y(x) = \frac{1}{(b-ax)}
f(z) = \frac{1}{(z)}
z(x) = b-ax
f'(z) = \frac{1}{z^2}
z'(x) = -a
y'(x) = \frac{-a}{(b-ax)^2}
\)
Could someone shed some light on this problem. It has to do with whether or not b-ax is seen as a composite function which I do not believe it is and as such the first derivation would be correct but if it is a composite function then the chain rule needs to be applied.
I am not using the quotient rule since the top part is not a function of x.
The following function is given:
\(
y(x) = \frac{1}{(b-ax)}
\)
I want to find the derivative of this one but I am stuck between two methods:
\(
y(x) = \frac{1}{(b-ax)}
y(x) = (b-ax)^{-1}
y(x) = b^{-1}-a^{-1}x^{-1}
y'(x) = 0 - a^{-1}*-x^{-2}
y'(x) = \frac{1}{a}*\frac{1}{x^{2}}
y'(x) = \frac{1}{ax^{2}}
\)
On the otherhand using the chain rule:
\(
y(x) = \frac{1}{(b-ax)}
f(z) = \frac{1}{(z)}
z(x) = b-ax
f'(z) = \frac{1}{z^2}
z'(x) = -a
y'(x) = \frac{-a}{(b-ax)^2}
\)
Could someone shed some light on this problem. It has to do with whether or not b-ax is seen as a composite function which I do not believe it is and as such the first derivation would be correct but if it is a composite function then the chain rule needs to be applied.
I am not using the quotient rule since the top part is not a function of x.