Evaluate the following derivatives of trigonometric functions:
1. f(x) = x² csc(2x)
2. f(x) = ³√cos²(2x+1)
with step-by-step process po sana, need it badly thank you!
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Evaluate the following derivatives of trigonometric functions:
1. f(x) = x² csc(2x)
2. f(x) = ³√cos²(2x+1)
with step-by-step process po sana, need it badly thank you!
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Answer:
Sure, here's how to evaluate the derivatives of these trigonometric functions:
1. f(x) = x² csc(2x)
Using the product rule, we can differentiate f(x) as follows:
f'(x) = (x²)' csc(2x) + x² (csc(2x))'
Now we just need to find the derivatives of each term. Recall that the derivative of x² is 2x, and the derivative of csc(x) is -csc(x) cot(x), so:
f'(x) = 2x csc(2x) - x² csc(2x) cot(2x)
Therefore, the derivative of f(x) is:
f'(x) = (2x csc(2x)) - (x² csc(2x) cot(2x))
2. f(x) = ³√cos²(2x+1)
Using the chain rule, we can differentiate f(x) as follows:
f'(x) = (³√cos²(2x+1))'
Let u = cos(2x+1), so f(x) = ³√u². Now we can apply the chain rule:
f'(x) = (³√u²)' du/dx
Recall that the derivative of u = cos(2x+1) is:
du/dx = -sin(2x+1) * 2
So we can substitute this in to get:
f'(x) = (³√u²)' * (-2 sin(2x+1))
Now we just need to find the derivative of (³√u²). We can write this as:
(³√u²) = u^(2/3)
Then using the power rule, we get:
(³√u²)' = (2/3)u^(-1/3) du/dx
Substituting this in, we get:
f'(x) = (2/3)u^(-1/3) * (-2 sin(2x+1))
Simplifying, we get:
f'(x) = (-4/3)cos(2x+1)^(-1/3) sin(2x+1)
Therefore, the derivative of f(x) is:
f'(x) = (-4/3)cos(2x+1)^(-1/3) sin(2x+1)