# Simple Example of Derivatives

## Implicit Differentiation (An example)

Q Differentiate ${\displaystyle y=arccos(2t/(1+t^{2}))}$

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

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 ${\displaystyle (x,y)}$ and ${\displaystyle (x+h,y+h)}$ and then take limit ${\displaystyle h->0}$

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

A ${\displaystyle f'(x)=lim(h->0)1/h*(x+h/(1+2(x+h)-x/(1+2x))=lim(h->0)h/h*(1+2x)(1+2(x+h)=1/(1+2x)^{2}}$

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

A slope of tangent = f'(x), equation of tangent line is ${\displaystyle y-f(x)=f'(x)(x-a)}$ [using the formula ${\displaystyle y-y1=m(x-x1)}$]

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

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

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

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

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

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

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