Find the derivative of (ax+b)^m / (cx+d)^n with respect to x

Steps to Solve

1. Apply the quotient rule

The quotient rule states that if we have a function $y = \frac{u}{v}$, then its derivative is given by $\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$. Here, let $u = (ax+b)^m$ and $v = (cx+d)^n$.

2. Find $\frac{du}{dx}$

Using the chain rule, $\frac{du}{dx} = m(ax+b)^{m-1} \cdot a = am(ax+b)^{m-1}$.

3. Find $\frac{dv}{dx}$

Similarly, using the chain rule, $\frac{dv}{dx} = n(cx+d)^{n-1} \cdot c = cn(cx+d)^{n-1}$.

4. Substitute into the quotient rule formula

Substituting $u$, $v$, $\frac{du}{dx}$, and $\frac{dv}{dx}$ into the quotient rule formula: $$ \frac{dy}{dx} = \frac{(cx+d)^n \cdot am(ax+b)^{m-1} - (ax+b)^m \cdot cn(cx+d)^{n-1}}{[(cx+d)^n]^2} $$

5. Simplify the expression

Simplify the denominator: $$ \frac{dy}{dx} = \frac{(cx+d)^n \cdot am(ax+b)^{m-1} - (ax+b)^m \cdot cn(cx+d)^{n-1}}{(cx+d)^{2n}} $$

6. Factor out common terms

Factor out $(ax+b)^{m-1}(cx+d)^{n-1}$ from the numerator: $$ \frac{dy}{dx} = \frac{(ax+b)^{m-1}(cx+d)^{n-1}[am(cx+d) - cn(ax+b)]}{(cx+d)^{2n}} $$

7. Further simplification

Simplify by canceling $(cx+d)^{n-1}$ from numerator and denominator: $$ \frac{dy}{dx} = \frac{(ax+b)^{m-1}[am(cx+d) - cn(ax+b)]}{(cx+d)^{n+1}} $$

8. Expand the terms in the square brackets $$ \frac{dy}{dx} = \frac{(ax+b)^{m-1}[amcx+amd - cnax-cnb]}{(cx+d)^{n+1}} $$

9. Simplify the expression in the brackets $$ \frac{dy}{dx} = \frac{(ax+b)^{m-1}[(amc-acn)x+amd-cnb]}{(cx+d)^{n+1}} $$