Please help! For the function f(x) = 2x^6 + 3x^4 - 4x^3 + (1/x) - sin(2x) find the first four derivatives.

See the explanation

Step-by-step explanation:

We know that

f(x) = 2x⁶ + 3x⁴ - 4x³ + (1/x) - sin2x

Lets calculate the derivatives:

f'(x) = 6(2x⁵) + 4(3x³) - 3(4x²) -( 1/x²) - 2(cos2x)

f'(x) = 12x⁵ + 12x³ - 12x² - (1/x²) - 2cos2x

Similarly:

f''(x) = 60x⁴ + 36x² - 24x + (2/x³) + 4sin2x

f'''(x) = 240x³ + 72x - 24 - (6/x⁴) + 8cos2x

Rearrange:

f'''(x) - 240x³ +72x - (6/x⁴) + 8cos2x - 24

f(x) = 720x² + 72 + (24/x⁵) - 16sin2x

Rearrange:

f(x) = 720x² + (24/x⁵) - 16sin2x +72

37/3

step-by-step explanation:

step 1: simplify both sides of the equation.

7x+8(x+

1

4

)=3(6x−9)−8

7x+(8)(x)+(8)(

1

4

)=(3)(6x)+(3)(−9)+−8(distribute)

7x+8x+2=18x+−27+−8

(7x+8x)+(2)=(18x)+(−27+−8)(combine like terms)

15x+2=18x+−35

15x+2=18x−35

tep 2: subtract 18x from both sides.

15x+2−18x=18x−35−18x

−3x+2=−35

step 3: subtract 2 from both sides.

−3x+2−2=−35−2

−3x=−37

step 4: divide both sides by -3.

−3x

−3

=

−37

−3

x=

37

3

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step-by-step explanation:

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