~~equation of the tangent line at x=1 for f(x)=2sqrtx/4x^2-5
asked Jul 21, 2015 in Calculus Answers by anonymous

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First we need to differentiate f(x) wrt x. Put u=2sqrt(x)=2x^(1/2) then du=x^-(1/2)dx; put v=4x^2-5, then dv=8x. Apply d(u/v)=(vdu-udv)/v^2 or du/v-udv/v^2: f'(x)=(x^-(1/2)(4x^2-5)-2x^(1/2)*8x)/(4x^2-5)^2.

Now substitute x=1: f'(1)=(-1-16)=-17. This is the gradient of the tangent at x=1 and it's also the gradient of the tangent line, represented now by g(x)=-17x+a where a is a constant. Now to find a.

f(1)=2/(-1)=-2. The point (1,-2) must also be on the line g(x), so g(x)=-2=-17+a, so a=15 and g(x)=15-17x is the equation of the tangent line.

answered Jul 22, 2015 by Rod Top Rated User (487,100 points)
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