Asked by eyyy
Find the derivative of the following function:
1. f(x)=cos[cot(sec x^2)]
2. y=[(x+1)(x+2)/(x-1)(x-2) ]^n
3. D_x [4 cos(sin3x)]
4. D_x ((2x^2+1)/(3x^3+1))^2
5. D_x cotxy+xy=y sinx
1. f(x)=cos[cot(sec x^2)]
2. y=[(x+1)(x+2)/(x-1)(x-2) ]^n
3. D_x [4 cos(sin3x)]
4. D_x ((2x^2+1)/(3x^3+1))^2
5. D_x cotxy+xy=y sinx
Answers
Answered by
oobleck
just apply the chain and product/quotient rules. Then get creative with your algebra and trig.
1) df/dx = -sin[cot(sec x^2)]*(-csc^2(sec x^2))*(sec x^2 * tan x^2)*(2x)
2) dy/dx = n[(x+1)(x+2)/(x-1)(x-2) ]^(n-1) * 6(2-x^2)/((x-1)(x-2))^2
3. D_x = -4sin(sin 3x))*(cos 3x)(3)
4. D_x =2*((2x^2+1)/(3x^3+1)) * -(2x)(2x^2+1)(6x^3+9x-4)/((3x^3+1))^3
5. If you want to find dy/dx then use implicit differentiation.
-csc^2(xy)(y+xy') + y+xy' = y' sinx + y cosx
y' = -(y(cot^2(xy)+cosx)/(x csc^2(xy) - x + sinx)
If you just want to take the partial derivative (which is a bit odd, since you don't have a function of a and y), then I'd guess
f(x,y) = cotxy+xy - y sinx = 0
∂f/∂x = -y csc^2(xy) + x - y cosx
1) df/dx = -sin[cot(sec x^2)]*(-csc^2(sec x^2))*(sec x^2 * tan x^2)*(2x)
2) dy/dx = n[(x+1)(x+2)/(x-1)(x-2) ]^(n-1) * 6(2-x^2)/((x-1)(x-2))^2
3. D_x = -4sin(sin 3x))*(cos 3x)(3)
4. D_x =2*((2x^2+1)/(3x^3+1)) * -(2x)(2x^2+1)(6x^3+9x-4)/((3x^3+1))^3
5. If you want to find dy/dx then use implicit differentiation.
-csc^2(xy)(y+xy') + y+xy' = y' sinx + y cosx
y' = -(y(cot^2(xy)+cosx)/(x csc^2(xy) - x + sinx)
If you just want to take the partial derivative (which is a bit odd, since you don't have a function of a and y), then I'd guess
f(x,y) = cotxy+xy - y sinx = 0
∂f/∂x = -y csc^2(xy) + x - y cosx
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