Which number's derivative is SiNx to the fourth power

Which number's derivative is SiNx to the fourth power

∵∫(sinx)^4dx=∫[(1-cos(2x))/2]²dx
=∫[(1-2cos(2x)+cos²(2x))/4]dx
=∫[(1-2cos(2x)+(1+cos(4x))/2)/4]dx
=∫[3/8-cos(2x)/2+cos(4x)/8]dx
=3x / 8-sin (2x) / 4 + sin (4x) / 32 + C (C is the integral constant)
The derivative of 3x / 8-sin (2x) / 4 + sin (4x) / 32 + C (C is the integral constant) is the fourth power of SiNx

Find the derivative of SiNx? + sin? (3x)

Let SiNx = t
So SiNx? + sin? (3x) = t
　　sin3x=sin(x+2x)=3sinx-4sin³x=3t-4t³
　　sinx³+sin³(3x)=3t-4t²+t³=3t-3t³
Derivative of composite function (3t-3t 3) '= 3cosx-12sinxcosx = 3cos-6sin2x
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What's the derivative of sin to the fifth power

It's an indefinite integral
∫sinx^5dx
=-∫sinx^4dcosx
=-∫[(1-cosx^2]^2dcosx
=-∫(1-2cosx^2+cosx^4)dcosx
=-∫1dcosx+∫2cosx^2dcosx-∫cosx^4dcosx
=-cosx+2/3cosx^3-1/5cox^5
Checking results
Derivation of - cosx + 2 / 3cosx ^ 3-1 / 5cox ^ 5
=(-cosx)'+(2/3cosx^3)'-(1/5cox^5)'
=sinx+2/3*3*cosx^2(cosx)'-1/5*5cosx^4(cosx)'
=sinx-2cosx^2sinx+cosx^4sinx
=sinx(1-2cosx^2+cosx^4)
=sinx(1-cosx^2)^2
=sinx(sinx^2)^2
=sinx^5

Find the derivative of sin (3x + 1) to the third power (including sine)

f(x)=sin^3(3x+1)
Let sin (3x + 1) = u, 3x + 1 = v
Then:
f'(x)=[y'|u]*[u'|v]*[v'|x]
=[(u^3)']*[(sinv)']*[(3x+1)']
=[3u^2]*[cosv]*[3]
=[3sin^2(3x+1)]*[cos(3x+1)]*3
=9sin^2(3x+1)cos(3x+1)

Find the derivative of function: y = SiNx + A's x power + 1n5

cosx+lna*a^x

Find the derivatives of the following functions: (1) y = 1 + X / 2-x (2) y = 1 + 2 / x + 3 / x 2 + 4 / X to the power of (3) y = SiNx? 2 / x + 2 / X

1.y'=1/2-1=-1/2
2. Y '= - 2 / x? - 8 / x? - 12 / X to the fourth power
3.y'=(2xcosx²-sinx²)/x²=2cosx²-sinx²/x²

Logarithmic derivative of y = SiNx ^ cosx + cosx ^ SiNx

(SiNx ^ cosx + cosx + cosx ^ SiNx) has been widely used in many countries, such as SiNx ^ cosx + cosx ^ SiNx ^ SiNx ^ SiNx ^ SiNx ^ SiNx ^ SiNx ^ SiNx ^ SiNx / SiNx ^ SiNx / SiNx ^ SiNx / SiNx ^ SiNx ^ cosx + cosx + cos ^ 2x / SiNx, Y1 '= (- sinxlsinx + cos ^ 2x / SiNx) * SiNx ^ cosx is the same reason Y2' = (cosxlncosx-sin ^ 2x / cosx / cos x) * cosx x x * cosx * cosx * cosx * cosx * cosx * cosx * cosx * cosx * cosx * cosx * cosx * cosx * cosx * cosx * cos x * x * x * x * x, ^ SiNx so y '= (- sinxlnsinx + cos ^ 2x / SiNx) * SiNx ^ cosx + (cosxlncosx sin ^ 2x / cosx) * cosx ^ SiNx

Derivative y = (cosx) ^ SiNx by logarithm

lny=ln(cosx)^sinx=sinxlncosx y^'/y=cosxlncosx-sinx^2/cosx y^'=(cosxlncosx-sinx^2/cosx)*y=
(cosx)^sinx *cosxlncosx-(cosx)^sinx*sinx^2/cosx

Derivation, y = (SiNx) ^ cosx

Click to enlarge, then click to enlarge:

Y = (1-sinx) / (1 + cosx) derivation

Using the division algorithm of derivation, (F / g) '= (f' * G-F * g ') / G ^ 2, the derivative in the title is y' = [(1-sinx) '* (1 + cosx) - (1-sinx) * (1 + cosx)'] / (1 + cosx) ^ 2 = [- cosx * (1 + cosx) - (1-sinx) * (- SiNx)] / (1 + cosx) ^ 2 = (- 1-cosx + SiNx) / (1 + cosx) ^ 2