# Solving a Derivative Equation Faster

How can I solve this problem as fast as possible just using a piece of paper and a pen:

$f(x)=(3x+4)^4\\f''(x)=?\\$
Please show me how to do that.

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\begin{align*} f'(x)&=4(3x+4)^3\cdot 3\\ &=12(3x+4)^3\\ f''(x)&=12\cdot 3(3x+4)^2\cdot 3\\ &=108(3x+4)^2 \end{align*}

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I am assuming you know chain rule and exponentiation rules. I.e., $(f^n)'=nf^{n-1}f'$ – Daniel Montealegre Mar 13 '12 at 6:45
"Chain rule", right... I remember it, yes. I also tried this simple, basic rule: (U^n)'=U'.n.U^(n-1) . I like this form and I've encountered this kind of problems very much recently... Thank you, Daniel. – Kerim Atasoy Mar 13 '12 at 6:57

Induction generalizes these sorts of problems; if you can intuitively remember these two formulas:

$$\frac{d^k}{dx^k} f(ax+b)=a^k \frac{d^kf}{dx^k}(ax+b);$$

$$\frac{d^k}{dx^k} x^n=n(n-1)\cdots\big(n-(k-1)\big) x^{n-k},$$

then you're golden. The first just says that in order to interchange differentiation and $x\mapsto ax+b$ transformations, you just have to bring the scale factor outside to the appropriate power. The second says that the power is reduced by $k$, and you put a "falling factorial" in front of it that stops just before $n-k$.

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Thank you very much, anon. I think this might work also... :) – Kerim Atasoy Mar 13 '12 at 7:01