Find the differentiation of the function y arcsinx

1/sqrt(1-x^2)dxi.e. (arcsinx).'

1/siny)'

1/cosy=1/sqrt((1-sin^2(y)))1/sqrt(1-x^2)

sqrt is the open square root.

Common Differential Formulas:

1. y=c (c is a constant) y'=0

2、y=x^n y'=nx^(n-1)

3、y=a^x y'=a^xlna，y=e^x y'=e^x4、y=logax y'=logae/x，y=lnx y'=1/x5、y=sinx y'=cosx

6、y=cosx y'=-sinx

7、y=tanx y'=1/cos^2x

8、y=cotx y'=-1/sin^2x

The derivative of the function is equal to the reciprocal of the inverse derivative x=siny, i.e. (arcsinx).'

1/siny)'

1/cosy=1/sqrt((1-sin^2(y)))1/sqrt(1-x^2)

sqrt is the open square root.

Yes (arcsinx).'=1 sqrt(1-x 2)Ah, this is the most basic differential formula, I don't know it???

Summary. According to the derivative of composite functions, the derivative of arcsinx is first found and then the derivative of 2x. Differentiation should be done with dy and dx.

Find the differentiation of y=arcsin2x.

Find the differentiation of y=arcsin2x.

Is the cube above x or above sin?

The original question was filmed and sent over.

On . You take a look. Derivatives are first found for arcsinx and then for 2x.

Good. According to the derivative of composite functions, the derivative of arcsinx is first found and then the derivative of 2x. Differentiation should be done with dy and dx.

The derivative of the number of letters, the sensitivity of lead is equal to the inverse function.

The reciprocal of the good number of the guide bridge x = siny, i.e. (arcsinx).'=1/siny)'=1/cosy=1/sqrt((1-sin^2(y)))1/sqrt(1-x^2)

sqrt is the open square root.

If the grandchild is y = arcsin(x 2)] 2, dy = 4x arcsin(x 2)dx (1-x 4).

If y = arcsin(2x 2), dy = 4xdx (1-4x 4).

If y = arcsin(2x)] 2, dy = 4arcsin(2x)dx (1-4x 2).

Problem : y=cosx+arcsinx,x=0, find the differential dy|x = 0 differential. f'(x)· x+o( x), where o( x) tends to 0 with x.

Hence the main part of the linear form of y dy=f'(x) x is the differentiation of y. [6] It can be seen that differentiation, as a kind of operation of the function-slippery jujube number, is consistent with the operation of the derivative (function) number.

The central idea of differentiation is infinite division. Differentiation is the linear major part of the amount of change in a function. One of the basic concepts of diosmine calculus.

Example 1 y= x , dy = dx

Example 2 y= cosx , dy = -sinxdx example 3 y= x 2 , dy = 2x dxy=cosx+arcsinx

The two sides are divided differentially.

dy=d(cosx+arcsinx)

Separate differentiation. dy = sinx+ 1 (1-x 2) ]dx is substituted for x=0

dy| x=0

sin0+ 1 (1-0 2) ]dxdx: result : y=cosx+arcsinx, dy|x=0 = dx

First, according to the chain rule, for f(x) =arcsin(x), there is:

f'(x) =1 / sqrt(1 - x^2)

Then, for g(x) =cos(x), there is:

g'(x) =sin(x)

Next, we can use the Hehui width difference formula to calculate y = g(x) +f(g(x)) cos(x) +arcsin(sin(x)), and get:

y' =g'(x) +f'(g(x)) g'(x)

sin(x) +1 / sqrt(1 - sin^2(x)) cos(x)

Since x=0, therefore:

sin(0) =0

cos(0) =1

Substituting these values into the above equation yields:

y'(0) =sin(0) +1 / sqrt(1 - sin^2(0)) cos(0)

0 + 1 / sqrt(1 - 0^2) ×1

Thus the differentiation of y = cos(x) + arcsin(sin(x)) when x=0 is y'(0) =1。

y'=2xarcsinx+x2 1 Nianhe (1-x 2).

The boy type is staring at.

dy=[2xarcsinx+x 2 rental(1-x 2)]dx

dy=[1 2(arcsinx)'s -1 2nd power (1 such as let(1-x to the 2nd power))+2arctanx(1 (1+x to the 2nd power))]dx