Confused as to why this is wrong. Derivative product & chain rule. Reddit r/calculus

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#calculus #bprpcalculus #apcalculus #tutorial #math

[beastboyviv]

Its good to see that you actually explained the product rule to the student before just solving the problem and actually explained where he did mistake. There should be more teachers like you who actually care about teaching instead of just solving problems.👍👍👍

[cyrusyeung]

Alternative solution:

Multiply the x into the cube root gives:
2x(x^4 + sin(3x))^(1/3)
= 2(x^7 + x^3 sin(3x))^(1/3)

Hence,
d/dx(2x(x^4 + sin(3x))^(1/3))
= d/dx(2(x^7 + x^3 sin(3x))^(1/3))
= 2/3 (x^7 + x^3 sin(3x))^(-2/3) (7x^6 + 3x^2 sin(3x) + 3x^3 cos(3x))

Then, we factor out a x^(-2) from the 1st parenthesis and factor a x^2 from the 2nd parenthesie gives:

2/3 (x^4 + sin(3x))^(-2/3) (7x^4 + 3sin(3x) + 3x cos(3x))

[tdchayes]

I like to factor out the "smallest" (most negative, in this case) exponent -- (x^4+sin(3x))^(-2/3). This gives you a result with two factors, which can be much better to deal with if you need to take a second derivative.

[juanavelinobobadillabravo]

Se acepte o no se acepte, creo que es una salvaje (will) explicación.
Una cosa por otra, solo por curiosidad! Mister Stevens, habla ud, Cantones y/o.Mandarin.
Valoria mucho su respuesta.

[Samiul_]

Just for fun can you differentiate this function [e^{xln²√(x^ln(1/x))}]²

[sohailansari]

You know the formulas of derivatives for operations on "x" with respect to x. That's why you use chain rule to take derivative of let's say, sin(3x) with respect to 3x and then 3x with respect to x