a) d/dx[x^3+4/x^2+x^(1/2)]=d/dx[x^3]+4d/dx[x^(-2)]+d/dx[x^(1/2)]ddx[x3+4x2+x12]=ddx[x3]+4ddx[x−2]+ddx[x12]
Using the power rule:
=3x^2-8x^-3+1/2x^(-1/2)=3x2−8x−3+12x−12
=3x^2-8/(x^3)+1/(2sqrtx)=3x2−8x3+12√x
b) d/dx[5(2x-1)^4]=5d/dx[(2x-1)^2]ddx[5(2x−1)4]=5ddx[(2x−1)2]
The chain rule states that:
dy/dx = dy/(du) (du)/dxdydx=dydududx
Let y=u^4y=u4 while u=2x-1u=2x−1
Using the power rule:
(dy)/(du)=4g^3, \quad (du)/(dx)=2
5dy/dx=dy/(du) (du)/dx=40(2x-1)^3
c) d/dx[(x^2+4]/sinx]=d/dx[x^2/sinx]+4d/dx[1/sinx]
Solve for d/dx[x^2/sinx] first.
Using the product rule:
d/dx[f(x)*g(x)]=f(x)(dg)/dx+g(x)(df)/dx
d/dx[x^2*1/sinx]=d/dx[x^2]*1/sinx+d/dx[1/sinx]*x^2
=(2x)/sinx+d/dx[1/sinx]*x^2
Solve for d/dx[1/sinx]
Using the chain rule, let:
y=1/u while u=sinx
dy/(du)=-1/u^2
(du)/dx=cosx
dy/(du) (du)/dx=-cosx/u^2
Sub back u=sinx
dy/dx=-cosx/sin^2x
We can also use the reciprocal rule to solve for 1/sinx
d/dx[1/f(x)]=-(df)/dx*1/f^2(x)
Finally, d/dx[x^2/sinx]=(2x)/sinx-(x^2cosx)/sin^2x
Now solve for part 2
d/dx[1/sinx] has already been solved
d/dx[1/sinx]=-cosx/sin^2x
Now, sub back in.
d/dx[x^2/sinx]+4d/dx[1/sinx]=(2x)/sinx-(x^2cosx)/sin^2x-cosx/sin^2x
