Find f'(x) for f(x)=cos^2(5x^3).

Mathematics

2 answers:

-BARSIC- [3]3 years ago

6 0

d/dx cos^2(5x^3)

= d/dx [cos(5x^3)]^2

= 2[cos(5x^3)]

= - 2[cos(5x^3)] * sin(5x^3)

= - 2[cos(5x^3)] * sin(5x^3) * 15x^2

= - 30[cos(5x^3)] * sin(5x^3) * x^2

Explanation:

d/dx x^n = nx^(n - 1)

d/dx cos x = - sin x

Chain rule:

d/dx f(g(...w(x))) = f’(g(...w(x))) * g’(...w(x)) * ... * w’(x)

artcher [175]3 years ago

3 0

Step-by-step explanation:

f(x) = cos²(5x³)

f(x) = (cos(5x³))²

Use chain rule to find the derivative:

f'(x) = 2 cos(5x³) × -sin(5x³) × 15x²

f'(x) = -30x² sin(5x³) cos(5x³)

If desired, use double angle formula to simplify:

f'(x) = -15x² sin(10x³)

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Factor the polynomial function over the complex numbers. f(x)=x^4-x^3-2x−4 Enter your answer in the box. f(x) = The answer is: (

Vladimir79 [104]

Answer:

Factors: $(x+i\sqrt2)(x-i\sqrt2)(x+1)(x-2)=0$

Step-by-step explanation:

We are given a polynomial:

$f(x)=x^4-x^3-2x-4$

We have to factor the given polynomial into its complex factors.

The factorization can be done as follows:

$f(x)=x^4-x^3-2x-4 = 0\\x^4-x^3-2x-4 = 0\\x^4-4-x^3-2x=0\\\text{Identity: }a^2-b^2 = (a+b)(a-b)\\(x^4-4)-(x^3+2x) = 0\\(x^2+2)(x^2-2)-x(x^2+2) = 0\\(x^2+2)(x^2-2-x) = 0\\(x^2+2)(x^2-x-2) = 0\\(x^2+2)(x^2-2x-+x-2) = 0\\(x^2+2)((x(x-2)+1(x-2))=0\\(x^2+2)(x+1)(x-2)=0\\\text{Identity: }a^2-b^2 = (a+b)(a-b)\\(x^2-(\sqrt{-2})^2)(x+1)(x-2)=0\\(x+i\sqrt2)(x-i\sqrt2)(x+1)(x-2)=0$