How to differentiate?

Question

$y=(x^4+2x-22)^2$

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Answer 1 · 2018-03-21T13:12:37

Use chain rule to give $dy/dx = 2(x^4 + 2x -22)*(4x^3 + 2)$

$u=x^4 + 2x - 22$
$y= u^2$
$dy = 2u du$
$du = 4x^3 + 2 dx$

$dy/dx = 2(x^4 + 2x -22)*(4x^3 + 2)$

Answer 2 · 2018-03-21T13:17:26

$2(x^4+2x-22)(4x^3+2)$

We use chain rule, which states that,

$(df)/dx=(df)/(du)*(du)/dx$

Let $u=x^4+2x-22$ , then $(du)/dx=4x^3+2$ . Also, then we have $f=u^2, (df)/(du)=2u$ .

Combining, we get:

$(df)/dx=2u*(4x^3+2)$

Now, we need to undo the substitution, and we get

$=2(x^4+2x-22)(4x^3+2)$