Find an equation of the tangent line to the graph of the function at the given point.\\ $y = \frac{x}{\sqrt{49 + x^2}}$, $(0, 0)$
Added by Taylor S.
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To find the derivative of y with respect to x, we can use the quotient rule. Let u = x and v = \sqrt(49+x^(2)). Then, the derivative of y with respect to x is given by: y' = (u'v - uv') / v^2 y' = (\sqrt(49+x^(2)) - x(1/2)(49+x^(2))^(-1/2)(2x)) / (49+x^(2)) y' = Show more…
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