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Apr 9, 2015 · Find the equation of the line that is tangent to the curve at the point $(0,\sqrt{\frac{\pi}{2}})$. Given your answer in slope-intercept form. I don't know how can I get the tangent line, without a given equation!!, this is part of cal1 classes.
May 22, 2020 · In order to find the equation of a tangent line to a given function at a given point, you need to consider what a tangent line is. In order for a line to be...
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Aug 4, 2018 · You could use infinitesimals... The slope of the tangent line is the instantaneous slope of the curve. So if we increase the value of the argument of a function by an infinitesimal amount, then the resulting change in the value of the function, divided by the infinitesimal will give the slope (modulo taking the standard part by discarding any remaining infinitesimals). For example, suppose we ...
Fortunately, the technique of implicit differentiation allows us to find the derivative of an implicitly defined function without ever solving for the function explicitly. The process of finding [latex]\frac{dy}{dx}[/latex] using implicit differentiation is described in the following problem-solving strategy.
After all, differentiating is finding the slope of the line it looks like (the tangent line to the function we are considering) No tangent line means no derivative. Also when the tangent line is straight vertical the derivative would be infinite and that is not good either. How and when does non-differentiability happen [at argument \(x\)]?
For example, find the equation of the tangent to the curve at the point (1, 3). Step 1. Differentiate the function of the curve. If , then. Step 2. Substitute the x-coordinate of the given point into this derivative to find the gradient, ‘m’ The gradient anywhere on the curve is found by the gradient function, . The tangent is at the point ...
The slope of this tangent line is f'(c) ( the derivative of the function f(x) at x=c). A secant line is one which intersects a curve at two points. Click this link for a detailed explanation on how calculus uses the properties of these two lines to define the derivative of a function at a point.
