What is the equation of the tangent line of #f(x)=xsqrt(x-1)# at #x=4#?

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Answer 1 · 2018-06-24T05:27:17

#y=5/sqrt3x-20/sqrt3#

#f'(x)=x*1/(2sqrt(x-1))+sqrt(x-1)#

simplify

#f'(x)=x/(2sqrt(x-1))+sqrt(x-1)#

#f'(x)=x/(2sqrt(x-1))+sqrt(x-1)*(2sqrt(x-1))/(2sqrt(x-1))#

#f'(x)=x/(2sqrt(x-1))+(2(x-1))/(2sqrt(x-1))#

#f'(x)=(x+2(x-1))/(2sqrt(x-1))#

expand the brackets

#f'(x)=(x+2x-2)/(2sqrt(x-1))#

#f'(x)=(3x-2)/(2sqrt(x-1))#

plug into point-slope formula

#y-y_1=f'(x_1)(x-x_1)#

remembering that the derivative finds the slope of a function so #m=f'(x_1)#

to find #y_1#

#f(4)=(4)sqrt((4)-1)#

#y_1=4sqrt(3)#

to find #f'(x_1)#

#f'(4)=(3(4)-2)/(2sqrt((4)-1)#

#f'(4)=(12-2)/(2sqrt(3))#

#f'(4)=10/(2sqrt(3))#

#f'(4)=5/sqrt(3)#

plug in

#y-4sqrt(3)=5/sqrt(3)(x-4)#

add #4sqrt(3)# to each side

#ycancel(-4sqrt(3)+4sqrt(3))=5/sqrt(3)(x-4)+4sqrt3#

expand brackets

#y=5/sqrt(3)x-5/sqrt3(4)#

#y=5/sqrt3x-20/sqrt3#