Find the derivatives of:
1. H(x)= sin2xcos2x
The answer given is 2cos4x. My question is, how in the world did they get that!? Shouldn't the answer at least contain the sin function, either negative or positive seeing as it's the derivative of cos? Also, which rules are involved in this problem? Product and chain rule?
2. f(x)= (cotx) / (sinx)
I ended up with a mess of an answer that no where near resembled the solution.
Any help would be greatly appreciated! :)
H(x) = sin2x cos2x
= 1/2 sin4x
That help?
Using the product rule, it's a bit more work, but you can get the same answer:
dH/dx = 2cos2x cos2x - 2sin2x sin2x
= 2(cos^2 2x - sin^2 2x)
= 2cos4x
Don't forget your trig just because you're taking calculus!
#2
df/dx = (u'v - uv')/v^2
= (-csc^2 * sin - cot * cos)/sin^2
= (-csc - cos^2/sin)/sin^2
= -1/sin^3 - cot^2 csc
= -csc(csc^2+cot^2)
or, getting rid of the pesky cot,
f = cos/sin^2
f' = (-sin^3 - 2cos^2*sin)/sin^4
= (-sin^2 - 2cos^2)/sin^3
= -(1+cos^2)/sin^3
= -csc(csc^2+cot^2)
To find the derivative of the given functions, we need to use the product rule and the chain rule.
1. H(x) = sin^2(x)cos^2(x)
We can rewrite this function as:
H(x) = (sin(x))^2 * (cos(x))^2
To find the derivative of H(x), we need to apply the product rule. The product rule states that if we have two functions u(x) and v(x), the derivative of their product is given by:
d/dx (u(x) * v(x)) = u'(x) * v(x) + u(x) * v'(x)
Applying this rule to H(x):
H'(x) = (2sin(x) * cos^2(x)) + (sin^2(x) * 2cos(x))
Simplifying the expression:
H'(x) = 2sin(x) * cos^2(x) + 2sin^2(x) * cos(x)
Now, we can use the trigonometric identity: sin^2(x) + cos^2(x) = 1 to simplify further.
Replace sin^2(x) as 1 - cos^2(x):
H'(x) = 2sin(x) * cos^2(x) + 2(1 - cos^2(x)) * cos(x)
Expanding and simplifying:
H'(x) = 2cos^3(x) - 2cos^3(x) + 2cos(x)
H'(x) = 2cos(x)
So, the derivative of H(x) is 2cos(x).
2. f(x) = (cot(x)) / (sin(x))
To find the derivative of f(x), we need to use the quotient rule. The quotient rule states that if we have two functions u(x) and v(x), the derivative of their quotient is given by:
d/dx (u(x) / v(x)) = [u'(x) * v(x) - u(x) * v'(x)] / (v(x))^2
Applying the quotient rule to f(x):
f'(x) = [(-cosec^2(x)) * sin(x) - cot(x) * cos(x)] / (sin(x))^2
Now, we can simplify the expression:
f'(x) = [-sin(x) * sin(x) - cot(x) * cos(x)] / (sin(x))^2
f'(x) = [-sin^2(x) - cot(x) * cos(x)] / (sin(x))^2
Using the trigonometric identity: 1 + cot^2(x) = cosec^2(x), we can rewrite cot(x) as cosec(x) * sin(x):
f'(x) = [-sin^2(x) - (cosec(x) * sin(x)) * cos(x)] / (sin(x))^2
Expanding and simplifying:
f'(x) = [-sin^2(x) - sin(x) * cos(x) * cosec(x)] / (sin(x))^2
f'(x) = [-sin(x) - sin(x) * cos(x) * cosec(x)] / (sin(x))^2
Further simplification is not possible as the derivative of f(x) is quite complex. However, you may verify your result by comparing it to the given solution or using a symbolic differentiation tool like Wolfram Alpha or calculus software.
I hope this helps! If you have any further questions, feel free to ask.