Differentiate a tangent expansion to obtain a secant-square series

The Teacher fills the first derivative steps left blank in the Student copy:

$\begin{aligned}&y^{\prime}\\ & =2\sec^2(2x)\\ & =2(1+\tan^2(2x))\\ & =2(1+y^2).\end{aligned}$

At zero, f(0)=tan0=0 and $f^{\prime}(0)=2$. Use radians for this trigonometric differentiation.

The complete source chain, including the Teacher intermediate steps, is

$y^{\prime\prime}=\frac d{dx}\{2(1+y^2)\}=2(2y)y^{\prime}=4yy^{\prime},\qquad f^{\prime\prime}(0)=0.$

$\begin{aligned}&y^{\prime\prime\prime}\\ & =\frac d{dx}(4yy^{\prime})\\ & =4yy^{\prime\prime}+4y^{\prime}y^{\prime}\\ & =4(y^{\prime})^2+4yy^{\prime\prime}.\end{aligned}$

At zero this is $f^{\prime\prime\prime}(0)=4(2)^2+4(0)(0)=16$, establishing the requested identity and its value.

The Teacher supplies the zero-value table and factorial substitution:

$\begin{aligned}&\tan2x\\ & =0+2x+\frac0{2!}x^2+\frac{16}{3!}x^3+\cdots\\ & =2x+\frac83x^3+\cdots.\end{aligned}$

The constant and quadratic coefficients vanish; the cubic coefficient is $16/3!=8/3$.

Both copies’ callout asks how $\sec^2 2x$ is related to $\tan2x$. The Teacher completes the deduction:

$\frac d{dx}(\tan2x)=\frac d{dx}\left(2x+\frac83x^3+\cdots\right),$

$2\sec^2(2x)=2+8x^2+\cdots,\qquad \boxed{\sec^2(2x)=1+4x^2+\cdots}.$

Divide by the chain-rule factor 2 after differentiating. Independently, termwise differentiation is justified inside the tangent power series’ convergence radius, here $|x|<\pi/4$; the next nonzero tangent term is of fifth degree, so the next nonzero secant-square term is fourth degree.