a) #d/dx[x^3+4/x^2+x^(1/2)]=d/dx[x^3]+4d/dx[x^(-2)]+d/dx[x^(1/2)]#
Using the power rule:
#=3x^2-8x^-3+1/2x^(-1/2)#
#=3x^2-8/(x^3)+1/(2sqrtx)#
b) #d/dx[5(2x-1)^4]=5d/dx[(2x-1)^2]#
The chain rule states that:
# dy/dx = dy/(du) (du)/dx#
Let #y=u^4# while #u=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#
