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

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Answer 1 · 2017-12-02T19:54:38

$f (x) = \frac{1 + 2 x^{\frac{1}{2}}}{1 + x^{\frac{3}{2}}}$ $f(x)= (1+2x^(1/2))/(1+x^(3/2))$

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

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

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

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

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

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

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

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

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))f(x)=1+2x121+x32f(x)= (1+2x^(1/2)) / (1+x^(3/2)) at (4, 5/9)?