How do you find the slope of a tangent line to the graph of the function #f(x)= (1+2x^(1/2)) / (1+x^(3/2))# at (4, 5/9)?

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Answer 1 · 2017-12-02T19:54:38

#f(x)= (1+2x^(1/2))/(1+x^(3/2))#

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

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

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

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

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

#f'(4)=(1/2-3/2*2-4)/(1+8)^2#

#f'(4)=(-13/2)/81=-13/162#

slope of a tangent line= #m=f'(4)= =-13/162#