Derivative of y = 3x^3 e^x sinx lnx

In this video, I showed how to apply the product rule to a product of multiple functions

[@dereklenzen2330]

You are a brilliant math teacher. When I saw this problem in the thumbnail, my first thought was "logarithmic differentiation." Instead, I discovered the generalized product rule for more than two functions for the first time, in spite of the fact that I have been doing calculus for years. Thanks! 😁

[@PrimeNewtons]

Thanks. Even though I am a big fan of logarithmic differentiation, I believe the generalized product rule is more efficient for strictly products. That's why I did it.

[@Alians0108]

If you do logarithmic differentations, any amount of products is trivial f(x) = g(x)*h(x)+.... ln(f(x)) = ln(g(x)*h(x)*...) = ln(g(x))+ln(h(x))+ln(...).+... f'(x)/f(x) = g'(x)/g(x)+h'(x)/h(x).... So regardless of how many products you have it's always gonna be f'(x) = f(x) * (g'(x)/g(x)+h'(x)/h(x)+....) which seems far easier ---- Quick example if f(x) = x^n*e^x*arcsin(x) then f'(x)/f(x) = nx^(n-1)/x^n + e^x/e^x + 1/sqrt(1-x^2)/arcsin(x) f'(x) is just that times f(x) Ex 2: f(x) = x^n = x*x*x*x*... n times f'(x) = x^n*(1/x+1/x+1/x+1/x.. n times) = n*x^n/x = nx^(n-1)

[@oraz.]

I thought you could use the product rule once for fg * hk and then do logarithmic differentiation but it's still complicated so then I watched the video which is a simpler way.

[@Samir-zb3xk]

Whenever theres more than 2 factors i would just do logarithmic differentiation lol y = (x³)(e^x)(sin(x))(ln(x)) ln(y) = 3ln(x) + x + ln(sin(x)) + ln(ln(x)) (1/y)(dy/dx) = 3/x + 1 + cot(x) + 1/(xln(x)) dy/dx = [3/x + 1 + cot(x) + 1/(xln(x))](x³)(e^x)(sin(x))(ln(x))

[@ThePayner11]

Exactly what I did! Just easier that way haha

[@adamkucera9094]

I like you log method better more good good

[@wafflaaar1067]

genius

[@samirayoub6100]

Thanks a lot, you make me able to understand things, I loosed their reasoning for more than 40 years. It's a pity I can't find applications where such formulas facilitate life.

[@johnroberts7529]

A bit of an organisational monster! An elegant result that I don't think I have seen before. I'm guessing that there may be a general result that could be proved by Induction. Thanks again Maestro. 😊

[@ThePayner11]

I did it in a completely different way 😅 f(x) = (x^3)*(e^x)*sin⁡(x)*ln⁡(x) → ln(f(x)) = ln⁡((x^3)*(e^x)*sin⁡(x)*ln⁡(x)) = ln⁡(x^3) + ln⁡(e^x) + ln⁡(sin⁡(x)) + ln⁡(ln⁡(x)) → d/dx (ln⁡(f(x))) = d/dx (3*ln⁡(x) + x + ln⁡(sin⁡(x)) + ln⁡(ln⁡(x))) → f'(x))/f(x) = 3/x + 1 + cot⁡(x) + 1/(x*ln⁡(x)) → f'(x))/f(x) = (1/(x*ln(x)*sin(x))) (3*sin⁡(x)*ln⁡(x) + x*sin⁡(x)*ln⁡(x) + x*ln⁡(x)*cos⁡(x) + sin⁡(x)) → f'(x) = ((x^2)*(e^x))*(3*sin⁡(x)*ln⁡(x) + x*sin⁡(x)*ln⁡(x) + x*cos⁡(x)*ln⁡(x) + sin⁡(x)) Note: For d/dx (ln(ln(x)), let y = ln(x): → dy/dx = 1/x → d/dy (ln(y)) = 1/y = 1/ln(x) → d/dx (ln(ln(x))) = (dy/dx)(d/dy) = (1/x)(1/ln(x)) = 1/(x ln(x)) For d/dx (ln(sin(x)), let u = sin(x): → du/dx = cos(x) → d/du (ln(u)) = 1/u = 1/sin(x) → d/dx (ln(sin(x)) = (du/dx)(d/du) = (cos(x))(1/sin(x)) = cot(x) Still got the same answer!

[@maxvangulik1988]

d/dx(x^3•e^x•sin(x)•ln(x)) =x^2•e^x•((x+3)sin(x)ln(x) +sin(x) +xcos(x)ln(x)) by the extended product rule

[@adw1z]

My thought was to use the 4-product rule: f’(x) = 3x^2 e^x sin(x) ln(x) + f(x) + x^3 e^x cos(x) ln(x) + x^2 e^x sin(x) Why? (fg)’ = f’g + fg’ (fgh)’ = (fg)’h + fgh’ = f’gh + fg’h + fgh’ Simply use induction to show that for functions ai: (a1a2…ak)’ = a1’a2…ak + a1a2’…ak + … + a1a2…ak’

[@eurocouto]

Great!... I didn't know this procedure: It is a surprising novelty!... Thanks so much, Newton!

[@sunillGD]

the case for derivative of product of n functions can probably be proven easily by induction, yeah? Though its probably faster to just derive it the regular way with just 4 functions

[@kinshuksinghania4289]

It's better to use logarithm on such functions and then proceed with differentiation. Logarithm turns the product into a sum and makes everything neat.

[@ryanmcauley1561]

This may be the first calculus video that I stopped and said out loud, "This is brilliant".

[@MathFromAlphaToOmega]

For an extra challenge, see if you can integrate it now. :P

[@surendrakverma555]

Very good explanation Sir. Thanks 🙏🙏🙏🙏🙏🙏

[@maburwanemokoena7117]

when the products are too many I prefer to just first take a log.

[@ericzacher509]

You made a mistake there you forgott to write the derivative of sin(x)

[@AnnaOkrutna-sd3ys]

For me integrals are far easier than derivatives, no idea why.

[@dean532]

The gradient under the influence of x 😅

[@MASHabibi-d2d]

Thanks for an other video master

[@joelmacinnes2391]

Although you wouldn't be penalised for missing this out, you could also take out sin(x) and cos(x) would go to cot(x)

[@gamaltabee6889]

We can take ln() for both sides this way we turn the product into sums Just differentiate 5 ln() and multiply by Y And you have the derivative and the factors are taken already

[@kennethgee2004]

well one could argue that you could factor out (x^2)(e^x)(sin x)(ln x). the sin x would become 1 and the cos x /sin x is cot x. all the ln x also become 1 except that the last term is 1/ ln x which is ln x ^-1 and power rule for log becomes - ln(x). it is still a mess and please do not ask us to graph this function or find all its zeros and thus the critical points.

[@colina64]