Pattern to ease differentiation.

Question

Is there a way to differentiate the following without multiplying everything out?

$$f(x)=\bigg(1 + \Big( 2 + \big(3 + (4 +x^6)^2~\big)^3~\Big)^4\bigg)^5$$

(Chain rule doesn't help much, binomial theorem neither.)

Answer 1

Chain rule really isn't that bad: \begin{align*} f'(x)&=5\left(1+\left(2+\left(3+\left(4+x^6\right)^2\right)^3\right)^4\right)^4\cdot 4\left(2+\left(3+\left(4+x^6\right)^2\right)^3\right)^3\\ &\quad\cdot 3\left(3+\left(4+x^6\right)^2\right)^2\cdot 2(4+x^6)\cdot 6x^5. \end{align*} Hopefully I didn't mess that up.

Answer 2

Hint: The chain rule does work. Here is a schematic way to look at it. Let $f_{a,k,m}(x) = (a + x^k)^m$. Then using the chain rule $f'_{a,k,m}(x) = mk(a + x^k)^{m-1} x^{k-1}$. Now using this notation, what you're looking for is the derivative of $$f(x) = f_{1,4,5} \circ f_{2, 3, 4} \circ f_{3,2,3} \circ f_{4,6,2}(x).$$ This will give you (by repeated use of the chain rule) $$ f'(x) = \\f_{1,4,5}'\circ f_{2, 3, 4} \circ f_{3,2,3} \circ f_{4,6,2}(x) \times f'_{2,3,4} \circ f_{3,2,3} \circ f_{4,6,2}(x) \times f'_{3,2,3}\circ f_{4,6,2}(x) \times f'_{4,6,2}(x). $$ Messy still, but much better than trying to expand the algebraic expression.

Answer 3

From inside outside:

$$6x^5\cdot2(4+x^6)\cdot3(3+(4+x^6)^2)^2\cdot\ldots$$