Question
Find an equation of the tangent line to the given curve at the given point.$$y=\sqrt{1+2 x^{2}} \text { at } x=2$$
Step 1
The derivative of the function $y=\sqrt{1+2x^{2}}$ is given by the chain rule as: \[y'=\frac{1}{2}(1+2x^{2})^{-\frac{1}{2}}\cdot(4x)\] Simplifying this, we get: \[y'=2x(1+2x^{2})^{-\frac{1}{2}}\] Show more…
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