Calculus 12

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Derivatives and their Applications

Implicit Differentiation

There will be times that you can't do normal differentiation, for example, if you are given the equation or a circle $x^2 + y^2 = r$. Sometimes, you can rearrange the equation, solve for y, and come up with a differential, but often not. Instead, we will take the derivative of both sides of the equation: $$ \frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}r$$ $$\frac{d}{dx}x^2 + \frac{d}{dx}y^2 = 0$$ $$2x + \frac{d}{dx}y^2 = 0$$ So, what do we do about the $y^2$? We can't take the derivative wrt x, since there is no x. So, just how you would multiply a fraction by $\frac{a}{a}$ to get it to a common denominator, we can do the equivalent with the derivative. $$ 2x + \frac{d}{dx}y^2 = 0$$ $$2x + \frac{dy}{dx}\frac{d}{dy}y^2 = 0$$ $$2x + \frac{dy}{dx}2y = 0$$ $$\frac{dy}{dx}2y = -2x$$ $$\frac{dy}{dx} = \frac{-2x}{2y}$$ $$\frac{dy}{dx} = -\frac{x}{y}$$ Looks bizarre, treating differentiation like a fraction, but it works. Just remember that all the rules of normal differentiation apply, so if you wind up with, for example $\frac{d}{dx}xy$, you need to apply the product rule to get $y\frac{d}{dx}x + x\frac{d}{dx}y$

What's really happening

We can do this because y is really a function of x, even if it is difficult or impossible to solve for y and take a "normal" derivative. So consider replacing all instances of y with g(x), differentiate by the normal rules, and then substitute back y for g(x) and $\frac{dy}{dx}$ for g'(x) $$ \frac{d}{dx}(x^2 + (g(x))^2) = \frac{d}{dx}r$$ $$\frac{d}{dx}x^2 + \frac{d}{dx}(g(x))^2 = 0$$ $$2x + 2g(x)g'(x) = 0$$ $$2x + 2y\frac{dy}{dx} = 0 $$

Common derivatives

$$ \frac{d}{dx}x^n = nx^{n-1}$$ $$\frac{d}{dx}A^x = ln(A)A^x$$ $$\frac{d}{dx}sin(x) = cos(x)$$ $$\frac{d}{dx}cos(x_ = -sin(x)$$ $$\frac{d}{dx}tan(x) = sec^2(x)$$ $$\frac{d}{dx}(f(x) \times g(x)) = f'(x) \times g(x) + g'(x) \times f(x) $$

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