Question

Derivada de y = ln[cos(x^3 - 1)^4]

Answer 1

y = ln[cos(x³ - 1)⁴]

u=x³-1

v=cos u

k=v⁴

p=ln (k)

y'= u' * v' * k' * p'

y'=(3x²) * (-sen u) * 4v³* 1/k

y'=(3x²) * (-sen (x³-1)) * 4cos³u * 1/v⁴

y'=(3x²) * (-sen(x³-1)) * 4cos³(x³-1) * 1/cos⁴u

y'=(3x²) * (-sen (x³-1)) * 4cos³(x³-1) * 1/cos⁴(x³-1)

y'=(3x²) * (-sen (x³-1)) * 4 * 1/cos(x³-1)

y'=(12x²) * (-sen (x³-1)) * 1/cos(x³-1)

y'=(-12x²) * tan (x³-1) = 12x²* tan (1-x³)

Answer 2

Acompanhe:

$\displaystyle y = \ln (\cos(x^3-1)^4) \\ \\ \\ y' = \frac{1}{\cos(x^3-1)^4} \cdot \cos(x^3-1)^4' \\ \\ \\ y' = \frac{1}{\cos(x^3-1)^4} \cdot 4 \cdot \cos(x^3-1)^{4-1} \cdot \cos'(x^3-1) \cdot (x^3-1)' \\ \\ \\ y' = \frac{1}{\cos(x^3-1)^4} \cdot4 \cdot \cos(x^3-1)^{3} \cdot -\sin(x^3-1) \cdot 3x^2 \\ \\ \\ y'=-\frac{12x^2 \cdot \cos(x^3-1)^3 \cdot \sin(x^3-1)}{\cos(x^3-1)^4} \\ \\ \\ y'=-\frac{12x^2 \cdot \cos(x^3-1)^0 \cdot \sin(x^3-1)}{\cos(x^3-1)^{4-3}}$

$\displaystyle \boxed{\boxed{ y'=-\frac{12x^2 \cdot \sin(x^3-1)}{\cos(x^3-1)} }}$