Differentiate only after keeping enough terms · source reference

Printed question

It is given that $y=\sqrt{1+e^{\tan x}}$.

Show that $2y\frac{dy}{dx}=(y^2-1)\sec^2x$, and hence show that when $x=0$, $\frac{d^2y}{dx^2}=\frac{3\sqrt2}{16}$. [5]

Find

(i) the first three terms of the Maclaurin’s series for $y$, [1]

(ii) the first two terms of the Maclaurin’s series for $\frac{dy}{dx}$, [2]

giving your answers in surd form.

Use the series to obtain an approximate value for $\int_{-1}^1 y\,dx$, giving your answer in surd form. [2]

[2009/MI/PU2 Promo/I/7]

Printed working

$y^2=1+e^{\tan x}$; $2yy'=\sec^2x\,e^{\tan x}=\sec^2x(y^2-1)$.

$2yy''+2(y')^2=(y^2-1)2\sec^2x\tan x+2yy'\sec^2x$.

At 0, $y=\sqrt2$, $y'=\sqrt2/4$; $2\sqrt2y''+2(\sqrt2/4)^2=0+2\sqrt2(\sqrt2/4)=1$; $y''=(3/4)/(2\sqrt2)=3\sqrt2/16$.

$y=\sqrt2+(\sqrt2/4)x+(3\sqrt2/16)x^2/2!=\sqrt2+\sqrt2x/4+3\sqrt2x^2/32+\cdots$.

$y'=\sqrt2/4+(3\sqrt2/32)(2x)+\cdots\approx\sqrt2(4+3x)/16$.

$\int_{-1}^1y\,dx\approx\sqrt2[x+x^2/8+x^3/32]_{-1}^1=\sqrt2(37/32+29/32)=33\sqrt2/16$.

Reading correction: the rendered question requests the first two terms of $y'$. A previous extracted-text reading said three; the printed two-term answer is complete for this request. The extra quadratic derivative term in Porisma’s earlier reviewed explanation is supplementary, not a missing source answer.

Teacher note 1

Square the equation so the resulting equation is easier to handle for repeated differentiation using implicit differentiation and product rule.

Use implicit differentiation and product rule:

$$\frac d{dx}(yy')=y\frac d{dx}(y')+y'\frac d{dx}(y)=yy''+(y')^2.$$

Also $\frac d{dx}e^{f(x)}=e^{f(x)}f'(x)$ and $\frac d{dx}\sec^2x=2\sec x(\sec x\tan x)$.

Teacher note 2

Let $y=f(x)$. To find the Maclaurin’s series for $y$, up to and including the term in $x^2$, differentiate up to the 2nd derivative. Substitute the values of $y,y',y''$ when $x=0$ into Maclaurin’s theorem:

$$f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3+\cdots.$$

Mathematical review

The printed two-term derivative answer is complete. The quadratic approximation is $y\approx\sqrt2(1+x/4+3x^2/32)$ and $y'\approx\sqrt2(4+3x)/16$. The symmetric integral cancels the odd term and gives $33\sqrt2/16$. The use of this truncation across the full interval $[-1,1]$ is an approximation, without a supplied error guarantee. A previous extracted-text reading falsely reported the question as requesting three derivative terms; the visually checked original says two.

Provenance

TJC JC2 2025 H2 Maths (9758) MSM. Native transcription checked against original visible question, working and sidebars. Question page(s): 22. Working page(s): 22, 23. The two Series source files are byte-identical; their shared visible content is expressed once and mapped to both page sets. Original records remain private for QA; this native article has no PDF or raster worksheet dependency.