Y = e ^ x cosx differential

Y = e ^ x cosx differential

dy=(e^xcosx-e^xsinx)dx=e^x(cosx-sinx)dx

Differential of y = cosx / 1-x ^ 2

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y=e^x*(sinx-cosx)
y'=e^x(sinx-cosx)+e^x(cosx+sinx)
=e^x(sinx+sinx)
=2sinxe^x.

Find the differential of y = e Λ x (SiNx cosx)

y'=dy/dx=e^x(sinx-cosx)+e^x(cosx+sinx)
dy=e^x(sinx-cosx)+e^x(cosx+sinx)dx

Find the differential of the following functions: ① y = 3x ^ 2 ② y = LNX ^ 2 ③ y = (e ^ - x) cosx

1) dy=6xdx
2) dy=1/x² *2xdx=2dx/x
3) dy=[-e^(-x)cosx-e^(-x)sinx]dx=-e^(-x)(cosx+sinx)dx

To find the differential of function y = xsin2x

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y = cosx/(1+sinx)dy/dx = [(1+sinx)(-sinx)-(cosx)(cosx)]/(1+sinx)²= (-sinx-sin²x-cos²x)/(1+sinx)²= [-sinx-(sin²x+cos²x)]/(1+sinx)²= -(sinx+1)/(1+sinx)²= -1/(1+si...

Approximate value by differential: √ (1.05), Tan 46 °, ln (1.002)

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Approximate value of Tan 46 ° by differential f(x)=tanx,f(x0)=tan45=1 =1+sec^(-2)*(π/4)×π/180=1+0.035=1.035 Why is f '(tan45) = sec ^ (- 2) * (π / 4)

For a function y = f (x), the differential at x = x0: dy = f '(x0) DX for this problem, y = f (x) = TaNx, f' (x) = sec 2x, x0 = π / 4, Δ x = DX = π / 180 (the angle should be converted into radian), then: the increment of function Δ y ≈ dy = f '(x0) DX = [sec 2 (π / 4)

In differential calculus of higher numbers, there is an approximate formula to find the approximate value, The third root of 128 requires an approximate formula to find the approximate value

Using the series expansion of F (x) = x ^ (1 / 3) at x = 125 (only one order approximation is needed)
f(125）=5
f'(125)=1/75
So f (x) = f (125) + (f '(125) / 1!) * (x-125) (approximate to the first order)
So f (128) = 5 + 1 / 75 * 3 = 5 + 1 / 25 = 5.04