how do i do i do y=cos(radt)+ rad(cost)?
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f(x) = (3x^2) (csc4x)^4 // f'(x)=?
my notes say : f'(x) = (3x^2)5(csc4x)^4(-csc(4x)(tan4x)(4)+(csc4x)^5(6)
I redid it and i got : f'(x) = (3x^2)5(csc4x)^4(-csc4x)(cot4x)(4) + (csc4x)^5(6x)
which is correct? if its my notes why?
To find the derivative of a function, you can use the rules of differentiation. Let's start with the first problem:
1. y = cos(rad(t)) + rad(cos(t))
To differentiate this expression, you can use the chain rule. We have two functions: cos(t) and rad(t).
- Differentiating cos(t) gives you -sin(t).
- Differentiating rad(t) with respect to t gives you 1/2 * (1/sqrt(t)).
Applying the chain rule, you multiply the derivatives of the inner and outer functions:
dy/dt = (-sin(t)) * (1/2 * (1/sqrt(t))) + (1/2 * (1/sqrt(cos(t)))) * (-sin(t))
Simplifying this expression may involve various trigonometric identities and algebraic manipulation, depending on the specific values of t involved.
Now let's move on to the second problem:
2. f(x) = (3x^2)(csc(4x))^4
Your notes show two different versions of the derivative. Let's break them down and compare:
Option 1 (as per your notes):
f'(x) = (3x^2) * 5 * (csc(4x))^4 * (-csc(4x) * tan(4x) * 4) + (csc(4x))^5 * 6
Option 2 (your version):
f'(x) = (3x^2) * 5 * (csc(4x))^4 * (-csc(4x) * cot(4x) * 4) + (csc(4x))^5 * (6x)
From a quick analysis, it seems that the difference lies in one term: -cot(4x) vs 6x.
To determine which version is correct, we would need to check the relevant trigonometric identities. Since I don't have access to the exact values of x or any specific instructions you're following, I cannot confirm which one is correct in your specific case.
However, you can check the identities yourself by using trigonometric rules like cotangent (cot(θ) = cos(θ)/sin(θ)) and seeing if it matches with either version. Additionally, you can try simplifying both versions algebraically to see if they are equivalent.
If you're following a specific textbook or class notes, it would be best to consult that resource for guidance on which version is correct. If in doubt, you can also reach out to your teacher or professor for clarification.