Answer:
f'(x) = [tex]\frac{-6((e^x - e^{-x})}{(e^x + e^{-x})^4}[/tex]
Step-by-step explanation:
Data provided in the question:
f(x) = [tex]\frac{2}{(e^x + e^{-x})^3}[/tex]
or
we can also write the above equation as:
f(x) = 2 × (eˣ + e⁻ˣ)⁻³
Now,
on differentiating the above equation with respect to 'x', we get
f'(x) = 2 × (-3) × (eˣ + e⁻ˣ)⁻⁴ × (eˣ + (-1)e⁻ˣ)
or
f'(x) = [tex]\frac{-6((e^x - e^{-x})}{(e^x + e^{-x})^4}[/tex]
Note: derivative of eˣ = eˣ
derivative of xⁿ = xⁿ⁻¹
