Question
Differentiate the following function with respect to x:
$(1-2\tan\text{x})(5+4\sin\text{x})$
$(1-2\tan\text{x})(5+4\sin\text{x})$
Then,
$\text{u}'=-2\sec^2\text{x};\text{v}'=4\cos\text{x}$Using the product rule:
$\frac{\text{d}}{\text{dx}}(\text{uv})=\text{uv}'+\text{vu}'$
$\frac{\text{d}}{\text{dx}}=[(1-2\tan\text{x})(5+4\sin\text{x})]$
$=(1-2\tan\text{x})(4\cos\text{x})+(5+4\sin\text{x})(-2\sec^2\text{x})$
$=4\cos\text{x}-8\times\frac{\sin\text{x}}{\cos\text{x}}\cos\text{x}-10\sec^2\text{x}-8\frac{\sin\text{x}}{\cos^2\text{x}}$
$=4\cos\text{x}-8\sin\text{x}-10\sec^2\text{x}-8\sec\text{x}\tan\text{x}$
$=4\Big(\cos\text{x}-2\sin\text{x}-\frac{5}{2}\sec^2\text{x}-2\sec\text{x}\tan\text{x}\Big)$
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