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Worked solution

Differentiate inverse trigonometric composites

Differentiate (a) cos⁡−1(2x)\cos^{-1}(2x), (b) tan⁡−1(x2−1)\tan^{-1}(x^2-1), and (c) sin⁡−1(ax+b)\sin^{-1}(ax+b) with respect to xx.

CJC · Lecture · Example 13

(a) Include the negative sign and inner derivative

The Teacher completion is −[1/1−(2x)2](2)=−2/1−4x2-[1/\sqrt{1-(2x)^2}](2)=-2/\sqrt{1-4x^2}. The real function is defined for −12≤x≤12-\frac12\le x\le\frac12; this finite derivative formula is valid for −12<x<12-\frac12<x<\frac12. Inverse cosine is not reciprocal cosine.

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(b) Differentiate the quadratic inside arctangent

The Teacher gives [1/(1+(x2−1)2)](2x)=2x/[1+(x2−1)2][1/(1+(x^2-1)^2)](2x)=2x/[1+(x^2-1)^2]. This is valid for all real xx, since the denominator is positive.

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(c) Treat a and b as constants

The printed chain-rule result is a/1−(ax+b)2a/\sqrt{1-(ax+b)^2}. For nonzero aa, the function needs ∣ax+b∣≤1|ax+b|\le1 and the displayed finite derivative applies where ∣ax+b∣<1|ax+b|<1. The endpoint-inclusive condition printed in the source rule table is a function-domain condition, not permission to divide by zero.

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(c) Check a constant inner function separately

If a=0a=0 and ∣b∣≤1|b|\le1, the function is the constant sin⁡−1b\sin^{-1}b and its derivative is 0. At b=±1b=\pm1 the printed quotient becomes 0/00/0, so use the constant-function argument instead. If ∣b∣>1|b|>1, there is no real-valued function. These parameter qualifications are independently authored.

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Common mistakes

The −1 in inverse-function notation is not a reciprocal exponent; retain the chain factor and check the denominator.

Solution notes

The Student copy leaves all three solutions blank. Complete blue Teacher working compared and integrated into the relevant parts. No printed marks or past-paper attribution. Endpoint and parameter qualifications are independently authored.

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