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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.

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Differentiate y=ln(x)+5x^2, and give the equation of the tangent at the point x=1

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