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Worked solution

Prove a squared-cosine derivative identity

Given y=32x+2sin⁡x+12sin⁡xcos⁡xy=\frac32x+2\sin x+\frac12\sin x\cos x, show that dy/dx=(1+cos⁡x)2dy/dx=(1+\cos x)^2.

CJC · Lecture · Self-Practice 2

Differentiate the product in the final term

dy/dx=32+2cos⁡x+12(cos⁡2x−sin⁡2x)dy/dx=\frac32+2\cos x+\frac12(\cos^2x-\sin^2x). The final term needs the product rule; the derivative of cos⁡x\cos x has a negative sign.

How did this part go?

Use an identity to complete the square

Substitute sin⁡2x=1−cos⁡2x\sin^2x=1-\cos^2x: dy/dx=32+2cos⁡x+12(2cos⁡2x−1)=1+2cos⁡x+cos⁡2x=(1+cos⁡x)2dy/dx=\frac32+2\cos x+\frac12(2\cos^2x-1)=1+2\cos x+\cos^2x=(1+\cos x)^2. This holds for all real xx in radians.

How did this part go?

Common mistakes

Differentiate the product sin x cos x before applying an identity.

Solution notes

Both copies print the requested result but no working for this Self-Practice. The intermediate proof here is independently authored.

Read the hint