How do you find the equation of the tangent line to the graph of #f(x) = 2x ln(x + 2) # at x = 3?

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Answer 1 · 2017-11-30T18:01:15

#y=(10ln(5)+6)/5x-18/5#

You first need to differentiate #f(x)# to find the gradient at #x=3#

#d/dx(a*b)= b*(da)/dx+a*(db)/dx#

Let #a=2x# and #b=ln(x+2)#

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

#->=2ln(x+2)+(2x)/(x+2)#

Gradient at #x=3#

#2ln(3+2)+(2(3))/((3)+2)=(10ln(5)+6)/5#

#(y-6ln(5))=(10ln(5)+6)/5(x-3)#

#y=(10ln(5)+6)/5x-(30ln(5)-18)/5+6ln(5)#

#->=(10ln(5)+6)/5x-18/5#

#:.#

#y=(10ln(5)+6)/5x-18/5#

Plot:

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