Question

How do you find the derivative of `f(x) = tan(sinx)`?

Answer 1

Let `y = tanu` and `u = sinx`. To apply the chain rule, we must differentiate both.

I'll start by showing you `u = sinx` using first principles so that you get an understanding of how this works.

`u' = lim_(h->0) (f(x + h) - f(x))/h`

Use the sum expansion formula `sin(A + B) = sinAcosB + cosAsinB`.

`u' = lim_(h->0) (sin(x + h) - sinx)/h`

`u' = lim_(h->0) (cosxsin h - sinx + sinxcos h)/h`

`u' = lim_(h->0) (cosxsin h - sinx(1 - cos h))/h`

`u' = " "cosx * lim_(h->0) sin h/h - sinx * lim_(h->0) (1 - cos h)/h`

We use the two well known limits `lim_(x-> 0) sinx/x = 1 and (1 - cosx)/x = 0` to evaluate.

`u' = cosx * 1 - sinx * 0`

`u' = cosx`

A similar process can be used to obtain the derivative of cosine.

Meanwhile, `y= tanu` can be written as `y = sinu/cosu`. This function, being of the form `f(x) = (g(x))/(h(x))`, can be differentiated using the quotient rule, where `color(magenta)(f'(x) = (g'(x) xx h(x)- g(x) xx h'(x))/(h(x))^2`.

You will find the derivative of `cosx` to be `-sinx`. Hence,

`y' = (cosu xx cosu - (sinu xx -sinu))/(cosu)^2`

Use the pythagorean identity `cos^2theta+ sin^2theta = 1`:

`y' = (cos^2u + sin^2u)/cos^2u`

`y' = 1/cos^2u`

Use the identity `1/costheta = sectheta`:

`y' = sec^2u`

Finally, we must determine the derivative of the entire function using the chain rule. This states that `color(magenta)(dy/dx = dy/(du) xx (du)/dx)`.

`dy/dx = sec^2u xx cosx`

`dy/dx = cosxsec^2(sinx)`

Hopefully this helps!