F(x)=x4-6x3+12x2-8, berapa turunan 3

A = x ( sec x + t an x ) 2 = 1 − s in x 1 + s in x ​ L = ( cos x 1 ​ + cos x s in x ​ ) 2 = ( cos x 1 + s in x ​ ) 2 = co s 2 x ( 1 + s in x ) 2 ​ = 1 − s i n 2 x ( 1 + s in x ) 2 ​ = 1 2 − s i n 2 x ( 1 + s in x ) 2 ​ = ( 1 − s in x ) ( 1 + s in x ) ( 1 + s in x ) 2 ​ = 1 − s in x 1 + s in x ​ = R
i f R = s in x 1 ​ 1 + s in x ​ = ( 1 + s in x ) s in x = s in x + s i n 2 x t h e n L  = R
sec x = cos x 1 ​ t an x = cos x s in x ​ s i n 2 x + co s 2 x = 1 → co s 2 x = 1 − s i n 2 x a 2 − b 2 = ( a − b ) ( a + b )

Answered by Anonymous | 2024-06-10

We have proven the identity ( sec A + tan A ) 2 = 1 − s i n A 1 + s i n A ​ by rewriting the left side using the definitions of secant and tangent, squaring the expression, and simplifying using the Pythagorean identity. Both sides of the equation match, confirming the identity holds true. Therefore, the proof is complete.
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Answered by Anonymous | 2024-10-02

Jawab:Penjelasan dengan langkah-langkah:The third derivative of the function f(x) = x⁴ - 6x³ + 12x² - 8 is f'''(x) = 24x - 36. Evaluating this at x=3 gives f'''(3) = 24(3) - 36 = 72 - 36 = 36.Here's how to find the third derivative:First derivative: f'(x) = 4x³ - 18x² + 24xSecond derivative: f''(x) = 12x² - 36x + 24Third derivative: f'''(x) = 24x - 36Now, substitute x = 3:f'''(3) = 24(3) - 36 = 72 - 36 = 36

Answered by sabarish1305 | 2025-07-07