Question 1192774
Use the chain rule,
{{{u=x+5}}}
{{{df/dx=(df/du)(du/dx)}}}
So,
{{{f=u^2}}}
{{{df/du=2u}}}
and
{{{u=x+5}}}
{{{du/dx=1}}}
So then,
{{{df/dx=(2u)(1)}}}
{{{df/dx=2(x+5)}}}
{{{df/dx=2x+10}}}
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.
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Similarly,
{{{v=x^2+2x+3}}}
{{{df/dx=(df/dv)(dv/dx)}}}
So,
{{{f=v^2}}}
{{{df/dv=2v}}}
and
{{{v=x^2+2x+3}}}
{{{dv/dx=2x+2}}}
So then,
{{{df/dx=(2v)(2x+2)}}}
{{{df/dx=2(x^2+2x+3)(2(x+1))}}}
{{{df/dx=4(x^2+2x+3)(x+1)}}}