Question #b5828

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Answer 1 · 2016-10-26T01:03:26

Ok, so we've got:

#y=x^2ln(2x)#

Now, let's transform this into:

#y=u*v#

Whereby, u is a function of x and v is a function of x .

If this is the case:

#(dy)/(dx)=u*(dv)/(dx)+v*(du)/(dx)#

This formula is actually called the product rule ** . Also, remember that y'** is the same as #(dy)/(dx)#.

Alright, so it turns out that:

#u=x^2# which means that, #(du)/(dx)=2x#.

Also:

#v=ln(2x)# which means that #(dv)/(dx)=1/x# thanks to the rule below:

When #y=ln(f(x))#, #(dy)/(dx)=(f'(x))/(f(x))#.

With the information we've just produced, we can now figure out what #(dy)/(dx)# will be...

#(dy)/(dx)=x^2*1/x+ln(2x)*2x#

#=x+2x*ln(2x)#

This result is y'*.

Now, using a similar strategy, let's figure out what #(d^2y)/dx^2# is. Remember that y'' is the same as #(d^2y)/dx^2#.

#(dy)/(dx)=x+2x*ln(2x)#

Which means that:

#(d^2y)/dx^2=1+d/(dx)(2x*ln(2x))#

#(d^2y)/dx^2=1+{2+2ln(2x)}#

#(d^2y)/dx^2=3+2ln(2x)#

This result is y''*.

If you'd like to know how the product rule can be produced from scratch, visit the link below. It may help you understand why we use it.