Wolfram Derivation Error

Question

I'm trying to derive the equation

$$y = (2x-6)^4$$

I thought that it would be

$$\frac{dy}{dx} = 8(2x-6)^3$$

Wolframalpha says $dy/dx = 64(x-3)^3$

Who's correct? I thought it would be a simple calc-1 chain rule.

Answer 1

$$[f(g(x))]'=g'(x)f'(g(x))$$ Taking $f(x)=x^4$ and $g(x)=2x-6$: $$[(2x-6)^4]'=(2x-6)'4(2x-6)^3=8(2x-6)^3=8(2(x-3))^3=64(x-3)^3$$

As you can see they are equivalent

Answer 2

$$\dfrac{dy}{dx}=8(2x-6)^3=8(2(x-3))^3=8\cdot(2^3(x-3)^3)=64(x-3)^3$$

As demostrated above, from the result you obtained to that obtained by Wolfram Alpha. As you can see, they are both equal.

Answer 3

There is no error. There is a common fact of $2$ in $(2x-6)$. That means we can write \begin{eqnarray*} (2x-6)^4 &\equiv& \left[2(x-3)\right]^4 \\ \\ &\equiv& 2^4\cdot(x-3)^4 \\ \\ &\equiv& 16(x-3)^4 \end{eqnarray*} If you differentiate this then you will get what $64(x-3)^3$ as Wolfram Alpha gives you.

Alternatively, if you differentiate first then you have \begin{eqnarray*} 8(2x-6)^3 &\equiv& 8\left[2(x-3)\right]^3 \\ \\ &\equiv& 8\cdot 2^3\cdot(x-3)^3 \\ \\ &\equiv& 64(x-3)^3 \end{eqnarray*}