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How to derive (ax+b)² ?
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Like, deriving the expanded form of this binomial?
I think that ; if we call U=x² and V=ax+b Drive like that : F'(x)=a*u'(V)
Ohh, derivatives. We could either differentiate term-by-term from the expanded form, or use chain rule. Chain rule: \( \displaystyle \frac{\text{d}}{\text{d}x} \left( (ax + b)^2 \right) \) \( \displaystyle = \frac{\text{d} \left(ax+b\right)^2}{\text{d}x} \) \( \displaystyle = \frac{\text{d} (ax + b)^2}{\text{d} (ax + b)} \times \frac{\text{d} (ax + b)}{\text{d}x} \)
d(ax+b)/dx = a If we call U=x^2 and V = ax + b, then.. d(ax+b)^2/d(ax+b) = d(U(V))/dV = U'(V) seems correct to me.
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