Find the derivatives of the following
18. y=〖cos〗^4 x^4
ANSWER: tanx2
19. y= sinx/(1+ 〖cos〗^2 x)
ANSWER: 3sinx
20. y=sinx(sinx+cosx)
ANSWER: sinx
Can you check my answers?
wow - remember the chain rule
if u = cos(x^4) then we have
y = u^4
y' = 4u^3 u'
y' = 4 cos^3(x^4) * (-sin(x^4)(4x^3)
= -16x^3 sin(x^4) cos^3(x^4)
y = sinx/(1+cos^2(x))
y' = [(cosx)(1+cos^2(x)) - (sinx)(2cosx)(-sinx)]/(1+cos^2(x))^2
y = sinx(sinx+cosx)
y' = cosx(sinx+cosx) + sinx(cosx-sinx)
I think you have some serious reviewing to do. I'd love to see the step-by-step work you did to get your answers.
-4 cos^3 x^4 sin x^4 (4 x^3)
= - 16 x^3 cos^3 x^4 sin x^4
how that is tan x2 whatever that is I can not imagine
ohhh.. I think I know what I did wrong.. I did not even realize that I was doing something completely different
Sure, I can help you check your answers.
18. To find the derivative of y = cos^4(x^4), we can use the chain rule. Let u = x^4, so y = cos^4(u).
Using the chain rule, we take the derivative of the outer function(cos^4) and multiply it by the derivative of the inner function(u).
Derivative of cos^4(u):
dy/dx = 4cos^3(u) * (-sin(u)) * du/dx
Since u = x^4, the derivative of u with respect to x (du/dx) is given by:
du/dx = 4x^3
Substituting these values back into the expression, we get:
dy/dx = 4cos^3(x^4) * (-sin(x^4)) * 4x^3
So, the correct answer for the derivative of y = cos^4(x^4) is:
dy/dx = -16x^3 * cos^3(x^4) * sin(x^4)
Therefore, your answer of "tanx2" is incorrect.
19. To find the derivative of y = sinx/(1+cos^2x), we can use the quotient rule.
Let u = sinx and v = 1+cos^2x.
Using the quotient rule, the derivative of y is given by:
dy/dx = (v * d(u/dx) - u * d(v/dx)) / v^2
To find the derivatives of u and v, we have:
d(u/dx) = d/dx(sin(x)) = cos(x)
d(v/dx) = d/dx(1+cos^2(x)) = -2cos(x)sin(x)
Substituting these values back into the expression, we get:
dy/dx = (1+cos^2x) * cos(x) - sin(x) * (-2cos(x)sin(x)) / (1+cos^2x)^2
Simplifying further:
dy/dx = cos(x) + cos^3(x) + 2sin^2(x)cos(x) / (1+cos^2x)^2
So, the correct answer for the derivative of y = sinx/(1+cos^2x) is:
dy/dx = cos(x) + cos^3(x) + 2sin^2(x)cos(x) / (1+cos^2x)^2
Therefore, your answer of "3sinx" is incorrect.
20. To find the derivative of y = sinx(sinx+cosx), we can use the product rule.
Let u = sinx and v = sinx+cosx.
Using the product rule, the derivative of y is given by:
dy/dx = u * d(v/dx) + v * d(u/dx)
To find the derivatives of u and v, we have:
d(u/dx) = d/dx(sin(x)) = cos(x)
d(v/dx) = d/dx(sinx+cosx) = cos(x)-sin(x)
Substituting these values back into the expression, we get:
dy/dx = sin(x) * (cos(x)-sin(x)) + (sin(x)+cos(x)) * cos(x)
Expanding and simplifying further:
dy/dx = cos(x)sin(x)-sin^2(x) + sin(x)cos(x) + cos^2(x)
Simplifying and combining like terms:
dy/dx = 2sin(x)cos(x)
So, the correct answer for the derivative of y = sinx(sinx+cosx) is:
dy/dx = 2sin(x)cos(x)
Therefore, your answer of "sinx" is incorrect.
I hope this helps you in understanding the correct derivatives and how to find them using the appropriate derivative rules.