Question

Find the dy/dx .... tant+3t , y =t/cost

Answer 1

Looks like
`x(t)=\tan t+3t` and
`y(t)=\frac t{\cos t}=t\sec t`.

By the chain rule,

`(\mathrm dy)/(\mathrm dx)=(\mathrm dy)/(\mathrm dt)(\mathrm dt)/(\mathrm dx)=((\mathrm dy)/(\mathrm dt))/((\mathrm dx)/(\mathrm dt))`

We have

`(\mathrm dy)/(\mathrm dt)=\sec t+t\sec t\tan t`

`(\mathrm dx)/(\mathrm dt)=\sec^2t+3`

so that the derivative is

`(\mathrm dy)/(\mathrm dx)=(\sec t+t\sec t\tan t)/(\sec^2t+3)`

We can simplify a bit. Multiply the numerator and denominator by
`\cos^2t`:

`(\mathrm dy)/(\mathrm dx)=(\cos t+t\sin t)/(1+3\cos^2t)`