Please help, thank you!!!
Find the derivative of the function.
g(θ) = (cos(8θ))9
If you meant g(θ) = (cos(8θ))^9 then
g ' (θ) = 9(cos(8θ))^8 * (-8sin(8θ))
is (cos(8x))^8 = (cos8x)^8
Some "purist" would claim that cos8x
would mean , take the cos8 then multiply that by x
to be safe, write it as cos(8x)
thank you so much :D
Could you please help me with this one too?? Thank you. Find d/dx for (sin πx + cos πy)4 = 65
To find the derivative of the function g(θ) = (cos(8θ))^9, we can use the chain rule. The chain rule states that if we have a function in the form (f(g(x)))^n, the derivative is given by:
d/dx (f(g(x))^n) = n(f(g(x)))^(n-1) * f'(g(x)) * g'(x)
In our case, f(u) = u^9, g(x) = cos(8x), and u = g(x). So, applying the chain rule, we get:
dg/dθ = 9((cos(8θ))^9-1) * (-sin(8θ)) * 8
Simplifying this expression, we have:
dg/dθ = -72sin(8θ)(cos(8θ))^8
Therefore, the derivative of g(θ) = (cos(8θ))^9 is -72sin(8θ)(cos(8θ))^8.