Question

How do you find the derivative of `y=tanx` using the limit definition?

Answer 1

`dy/dx = sec^2x`

Explanation:

By definition If `y=f(x)` then:
`dy/dx=f'(x) =lim_(hrarr0)( (f(x+h)-f(x))/h )`

So, with `y=tanx` we have:
`dy/dx = lim_(h rarr0)((tan(x+h)-tanx)/h)`

Using the trig identity for `tan (a + b)` we have;
`dy/dx = lim_(h rarr0)((((tanx+tan h)/(1-tanx*tan h))-tanx)/h)`

Putting over a common denominator of `1-tanxtan h`:
`:. dy/dx = lim_(h rarr0)((((tanx+tan h) - tanx(1-tanxtan h))/(1-tanxtan h))/h)`
`:. dy/dx = lim_(h rarr0)(((tanx+tan h-tanx+tan^2xtan h)/(1-tanxtan h))/h)`
`:. dy/dx = lim_(h rarr0)((tan h+tan^2xtan h)/(h(1-tanxtan h)))`
`:. dy/dx = lim_(h rarr0)((tan h(1+tan^2x))/(h(1-tanxtan h)))`
`:. dy/dx = lim_(h rarr0)((1+tan^2x)/(1-tanxtan h) * tan h/h)`
`:. dy/dx = lim_(h rarr0)((1+tan^2x)/(1-tanxtan h)) * lim_(h rarr0)(tan h/h)`

Now,
`lim_(hrarr0)(tan h/h)=lim_(hrarr0)(sin h/(cos h*h))`
`:. lim_(hrarr0)(tan h/h)=lim_(hrarr0)(sin h/h)*lim_(hrarr0)(1/cos h)`
`:. lim_(hrarr0)(tan h/h)=1*1` (standard trig limits)
`:. lim_(hrarr0)(tan h/h)=1`

And so,
`:. dy/dx= lim_(h rarr0)((1+tan^2x)/(1-tanxtan h))`
`:. dy/dx = (1+tan^2x)/(1-0)`
`:. dy/dx = 1+tan^2x`

And finally, using the standard trig identity `1+tan^2A-=sec^2A` we get the solution;
`dy/dx = sec^2x`