Differentiate y=ln(x)+5x^2, and give the equation of the tangent at the point x=1
First differentiate the equation, giving you, y'=(1/x)+10x. To get the gradient at this point of the curve, plug in x=1, to get a y' value of 11, and a y value of 5. From there you can plug these three numbers into the equation y-y1=y'(x-x1) to get the equation for the straight line y=11x-6.
HM
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