# Thread: differentiation of tant + tsec^2t

1. ## differentiation of tant + tsec^2t

Hello, I am getting confused as to how to differentiate this function:

tant + tsec^2t

I attempted it incorrectly as such:

sec^2t + (1)(sec^2t) + (t)(2sint/cos^3t)
by using product rule for the second part of the sum and using 2sinx/cos^3x as the differentiation of sec^2t

Can someone please show me where I went wrong ?

Thanks kindly

2. ## Re: differentiation of tant + tsec^2t

Originally Posted by fran1942
Hello, I am getting confused as to how to differentiate this function:

tant + tsec^2t

I attempted it incorrectly as such:

sec^2t + (1)(sec^2t) + (t)(2sint/cos^3t)
by using product rule for the second part of the sum and using 2sinx/cos^3x as the differentiation of sec^2t

Can someone please show me where I went wrong ?

Thanks kindly
\displaystyle \displaystyle \begin{align*} \sec^2{t} = \left(\cos{t}\right)^{-2} \end{align*}, so if we let \displaystyle \displaystyle \begin{align*} y = \left(\cos{t}\right)^{-2} \end{align*}, we can let \displaystyle \displaystyle \begin{align*} u = \cos{t} \implies y = u^{-2} \end{align*}, then

\displaystyle \displaystyle \begin{align*} \frac{du}{dt} &= -\sin{t} \\ \\ \frac{dy}{du} &= -2u^{-3} \\ &= -\frac{2}{\cos^3{t}} \\ \\ \frac{dy}{dt} &= \frac{2\sin{t}}{\cos^3{t}} \end{align*}

3. ## Re: differentiation of tant + tsec^2t

2sint/cos^3t is also equal to 2sec^2ttant. Perhaps they have used this in their answer.

4. ## Re: differentiation of tant + tsec^2t

It is worth memorising that the derivative of secx is secxtanx. Can then use chain rule for other expressions
Example For sec^2x (which means (secx)^2) we get 2secxsecxtanx=2sec^2xtanx