d/dx^2

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Why you dont reduce the fraction by d? Its much easier and faster

sinity

I could hear that marker saying: "AND YOU ARE DONE!"

justalittlechikon

Put x^2=t then proceed as usual .You will get the answer. DrRahul Rohtak Haryana India

dr.rahulgupta

I'll wash my eyes with bleach after seeing that

GoldenEyeMachit

Alternative way:
Let D=d/dx^2.
Then D(x^5+7)=D((x^2)^(5/2)+7)=5/2 (x^2)^(3/2)=5/2 x^3

nicolastorres

This is why Leibniz notation is confusing.

integralboi

Your channel should have a lot more of subscribers you deserve a lot

abdulrahimasfour

Integral ((x^6)/((x^12)+1)), please!!!

PhatNguyen-jmuz

Sir 1 doubt for me in complex numbers..
i²=-1
Now multiply the power with(2/2)
(i²)^2/2
Now ((i²)²)^½
Now (i⁴)^½
We know that i⁴ is 1..so (i⁴)^½ becomes 1^½ = +or- 1
But i²=-1 ...what is the reason for this incompatibility..? Please reply sir..

harrymed

Please give solution for:
integrate (sin x)^(cos x) with respect to dx

hari_g_k_._

This is really just the chain rule in disguise. Notice that dx^2/dx·df(x)/dx^2 = df(x)/dx, and dx^2/dx = 2·x, hence 2·x·df(x)/dx^2 = df(x)/dx. Therefore, df(x)/dx^2 = 1/(2·x)·df(x)/dx. In this video, f(x) = x^5 + 7, so df(x)/dx = 5·x^4, and 1/(2·x)·df(x)/dx = 5·x^4/(2·x), so df(x)/dx^2 = 5·x^4/(2·x), which is precisely the equality arrived at the end.

In general, dg(x)/dx·df(x)/dg(x) = df(x)/dx, so df(x)/dg(x) = [df(x)/dx]/[dg(x)/dx]. Notice that if g(x) = x, then this merely simplifies to the trivial df(x)/dx = df(x)/dx. Of course, this does assume that f and g are differentiable.

angelmendez-rivera