Simple Example of Derivatives

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Implicit Differentiation (An example)

Q Differentiate $y = a r c c o s (2 t / (1 + t^{2}))$

A A trick to do this question is to convert the question to $c o s (y) = 2 t / (1 + t^{2})$ . Now do implicit differentiation to get $- s i n (y) y^{'} = (- 2 t^{2} + 2) / (1 + t^{2})^{2}$ . Since as know $c o s (y) = 2 t / (1 + t^{2})$ , then $s i n (y) = (t^{2} - 1) / (t^{2} + 1)$ . (there're two ways to get this. One way is to use trig identity $s i n^{2} + c o s^{2} = 1$ ; the other way is to draw a right-angled triangle with angle $y$ , adjacent side $= 2 t$ and hypotenuse $= (1 + t^{2})$ . Then the opposite side is $(t^{2} - 1)$ , and $s i n (y) = o p p / h y p$ . So going back to the derivative and isolate $y^{'} = (- 2 t^{2} + 2) / (t^{4} - 1)$ .

Q Difference between secant and a tangent:

A A secant is any line passing through a graph. It would intersect the graph at least two points. However when two such points get closer and closer to each other, the secant becomes a tangent to a graph. One particular application is finding a derivative or slope of tangent at certain point. First find slope of the secant at $(x, y)$ and $(x + h, y + h)$ and then take limit $h - > 0$

Q Find the derivative of $f (x) = x / (1 + 2 x)$ using the definition of limits

A $f^{'} (x) = l i m (h - > 0) 1 / h * (x + h / (1 + 2 (x + h) - x / (1 + 2 x)) = l i m (h - > 0) h / h * (1 + 2 x) (1 + 2 (x + h) = 1 / (1 + 2 x)^{2}$

Q find the tangent to the graph f(x) at x=a

A slope of tangent = f'(x), equation of tangent line is $y - f (x) = f^{'} (x) (x - a)$ [using the formula $y - y 1 = m (x - x 1)$ ]

Q Find the derivative of $f (x) = 2^{e x} e^{2 x}$

A Using the product rule we can write $f^{'} (x) = 2^{e x} (e^{2 x})^{'} + (2^{e x})^{'} e^{2 x}$ .

Now the derivative of $e^{2 x}$ will be $2 e^{2 x}$ using chain rule

Derivative of $2^{e x} = e \log (2) 2^{e x}$ using chain rule

So $f^{'} (x) = 2^{e x} 2 e^{2 x} + e \log (2) 2^{e x} e^{2 x}$

Hint to find derivative of $2^{e x}$ : Derivative of $a^{x}$ is $a^{x} \log a$ using chain rule. use this formula $a^{x} = e^{x \log a}$ . An extra $e$ will come by derivative of $e x$

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