Sure! Let's differentiate the function [tex]\( f(x) = (ax + b)(cx + d) \)[/tex] step-by-step. We use the product rule for differentiation, which states that if we have a product of two functions [tex]\( u(x) \)[/tex] and [tex]\( v(x) \)[/tex], then
[tex]\[
\frac{d}{dx} [u(x) \cdot v(x)] = u'(x) \cdot v(x) + u(x) \cdot v'(x)
\][/tex]
In our case, let:
- [tex]\( u(x) = ax + b \)[/tex]
- [tex]\( v(x) = cx + d \)[/tex]
First, differentiate [tex]\( u(x) = ax + b \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[
\frac{d}{dx} (ax + b) = a
\][/tex]
Next, differentiate [tex]\( v(x) = cx + d \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[
\frac{d}{dx} (cx + d) = c
\][/tex]
Now, apply the product rule:
[tex]\[
\frac{d}{dx} [(ax + b)(cx + d)] = (ax + b) \cdot \frac{d}{dx} (cx + d) + (cx + d) \cdot \frac{d}{dx} (ax + b)
\][/tex]
Substitute the derivatives we found:
[tex]\[
\frac{d}{dx} [(ax + b)(cx + d)] = (ax + b) \cdot c + (cx + d) \cdot a
\][/tex]
Distribute the constants:
[tex]\[
\frac{d}{dx} [(ax + b)(cx + d)] = c(ax + b) + a(cx + d)
\][/tex]
Expand the expressions inside the parentheses:
[tex]\[
\frac{d}{dx} [(ax + b)(cx + d)] = acx + bc + acx + ad
\][/tex]
Combine like terms:
[tex]\[
\frac{d}{dx} [(ax + b)(cx + d)] = acx + ad + bcx + bc
\][/tex]
Group the like terms:
[tex]\[
\frac{d}{dx} [(ax + b)(cx + d)] = a(cx + d) + c(ax + b)
\][/tex]
Therefore, the derivative of the function [tex]\( f(x) = (ax + b)(cx + d) \)[/tex] is:
[tex]\[
\frac{d}{dx} \left[ (ax + b)(cx + d) \right] = a(cx + d) + c(ax + b)
\][/tex]
This result matches the given answer [tex]\( (a(cx + d) + c(ax + b), a(cx + d) + c(ax + b)) \)[/tex]. Hence, the simplified form of the derivative is indeed [tex]\( a(cx + d) + c(ax + b) \)[/tex].
