Differentiate only after keeping enough terms

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

Show that 2ydydx=(y2−1)sec⁡2x2y\frac{dy}{dx}=(y^2-1)\sec^2x, and hence show that when x=0x=0, d2ydx2=3216\frac{d^2y}{dx^2}=\frac{3\sqrt2}{16}. marks

Find

(i) the first three terms of the Maclaurin’s series for yy, mark

(ii) the first two terms of the Maclaurin’s series for dydx\frac{dy}{dx}, marks

giving your answers in surd form.

Use the series to obtain an approximate value for ∫−11y dx\int_{-1}^1 y\,dx, giving your answer in surd form. marks

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

Total marks: 10

Solution

Differential identity

Implicit differentiation

y=1+etan⁡xy=\sqrt{1+e^{\tan x}}. Show that 2ydydx=(y2−1)sec⁡2x2y\dfrac{dy}{dx}=(y^2-1)\sec^2x, and that d2ydx2=3216\dfrac{d^2y}{dx^2}=\dfrac{3\sqrt2}{16} at x=0x=0.

Hint

  1. Square: y2=1+etan⁡xy^2=1+e^{\tan x}.

  2. Differentiate twice and substitute x=0x=0.

Answer · Differential identity

2yy′=(y2−1)sec⁡2x2yy\prime=(y^2-1)\sec^2x; y′′(0)=32/16y\prime\prime(0)=3\sqrt2/16.

(i)

Maclaurin series

Find the first three terms of the Maclaurin series for yy.

Explore this idea

Hint

  1. y(0)=2y(0)=\sqrt2, y′(0)=24y'(0)=\frac{\sqrt2}4, y′′(0)=3216y''(0)=\frac{3\sqrt2}{16}.

Answer · (i)

2+2x/4+32x2/32+O(x3)\sqrt2+\sqrt2x/4+3\sqrt2x^2/32+O(x^3).

Explore the idea

y and its series

Compare the function with a finite polynomial. Change the highest included power and test x, then use the error plot to see how the approximation changes at that x.

Highest included powerxⁿ, n = 2
Test value xx = 0.60
Try an example
Function
1.72686
P₂(x)
1.67408
Error
-0.05279

The shading is a local comparison window, not a convergence or accuracy guarantee.

Teal marks the function, amber marks the polynomial, and the red segment joins their values at the selected x.

Does keeping more powers help at this x?

Highest included power n runs horizontally; signed error Pₙ(x) − f(x) runs vertically. Zero means agreement at this x. A missing power leaves the polynomial unchanged; individual steps need not improve the error.

  • Function
  • Series truncated at the chosen power
  • A local comparison window, not a convergence or accuracy guarantee
Key idea

The quadratic is accurate near 0 and drifts away towards ∣x∣=1|x|=1.

Ideas to try
Compare the polynomial with the function at x=1x=1.
(ii)

Series of the derivative

Find the first two terms of the Maclaurin series for dydx\dfrac{dy}{dx}.

Explore this idea

Hint

  1. Differentiate the series term by term.

Answer · (ii)

2/4+32x/16+O(x2)\sqrt2/4+3\sqrt2x/16+O(x^2).

Explore the idea

The derivative and its series

Compare the function with a finite polynomial. Change the highest included power and test x, then use the error plot to see how the approximation changes at that x.

Highest included powerxⁿ, n = 1
Test value xx = 0.50
Reset the controls
Function
0.67892
P₁(x)
0.48614
Error
-0.19278

The shading is a local comparison window, not a convergence or accuracy guarantee.

Teal marks the function, amber marks the polynomial, and the red segment joins their values at the selected x.

Does keeping more powers help at this x?

Highest included power n runs horizontally; signed error Pₙ(x) − f(x) runs vertically. Zero means agreement at this x. A missing power leaves the polynomial unchanged; individual steps need not improve the error.

  • Function
  • Series truncated at the chosen power
  • A local comparison window, not a convergence or accuracy guarantee
Key idea

The available quadratic series determines dy/dxdy/dx through degree 1, as requested. Higher derivative-series terms require more terms of the original function.

Ideas to try
Degree 1 is the tangent line to the derivative.
Integral

Integrating a series

Use the series to obtain an approximate value of ∫−11y dx\int_{-1}^1y\,dx in surd form.

Explore this idea

Hint

  1. Integrate the quadratic series from −1-1 to 11.

  2. Odd powers cancel.

Answer · Integral

∫−11y dx≈332/16\int_{-1}^1y\,dx\approx33\sqrt2/16.

Requested quadratic truncation; small error is not guaranteed over the entire interval. The source requests only two derivative terms.

Explore the idea

True area against the series estimate

Drag x to compare the numerical signed integral with the series estimate. Their difference shows the approximation error.

Upper limit xx = 1.00
Try an example

How the shaded areas add up

The limits run left to right. Area above the axis adds; area below the axis subtracts.

Above-axis contribution
2.979
Below-axis contribution
0
Signed integral
2.979

All three bars share one scale. Positive contributions extend right of zero; negative contributions extend left.

Signed integral from -1.00 to x
2.979
series antiderivative − F(−1)
2.917
Estimate − integral
-0.062
2.10% of the integral’s magnitude
Compare the running integral on a graph
  • Integrand
  • Numerical signed integral from the start to x
  • Part of the integrand above the axis
  • Part of the integrand below the axis
  • Series estimate of the signed integral; it need not agree with the true integral
Key idea

The odd term integrates to zero over [−1,1][-1,1], leaving 33216\frac{33\sqrt2}{16}.

Ideas to try
Drag xx to 1: compare the solid (true) area 2.979 with the dashed (series) value 2.917.