Differentiate in two ways, using product rule and otherwise, the function (1+2 )(5+4 ). verify that the answers are the same.

Let u=1+2 ;v=5+4 Then, u'=2 ^2x;v'=-4 Using the product rule: d dx (uv)=uv'+vu' d dx [(1+2 )(5+4 )] =(1+2 )(-4 )+(5+4 )(2 ^2x) =-4 -8 +10 ^2x+8 =-4 +10 ^2x+ (8 - 8 ^2x ) =-4 +10 ^2x+8 ( 1- ) =-4 +10 ^2x+8 ( ^2x ) =-4 +10 ^2x+8 Alternate Answer (1+2 )(5+4 )=5+4 +10 +8 Now we have, d dx [(1+2 )(5+4 )] = d dx (5+4 +10 +8 ) =-4 +10 ^2x+8

Differentiate in two ways, using product rule and otherwise, the …

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Question

Differentiate in two ways, using product rule and otherwise, the function $(1+2\tan\text{x})(5+4\cos\text{x}).$ verify that the answers are the same.

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Answer

Let $\text{u}=1+2\tan\text{x};\text{v}=5+4\cos\text{x}$Then, $\text{u}'=2\sec^2\text{x};\text{v}'=-4\sin\text{x}$
Using the product rule:
$\frac{\text{d}}{\text{dx}}(\text{uv})=\text{uv}'+\text{vu}'$
$\frac{\text{d}}{\text{dx}}[(1+2\tan\text{x})(5+4\cos\text{x})]\\=(1+2\tan\text{x})(-4\sin\text{x})+(5+4\cos\text{x})(2\sec^2\text{x})$
$=-4\sin\text{x}-8\tan\text{x}\sin\text{x}+10\sec^2\text{x}+8\sec\text{x}$
$=-4\sin\text{x}+10\sec^2\text{x}+\Big(\frac{8}{\cos\text{x}}-\frac{8\sin^2\text{x}}{\cos\text{x}}\Big)$
$=-4\sin\text{x}+10\sec^2\text{x}+8\Big(\frac{1-\sin\text{x}}{\cos\text{x}}\Big)$
$=-4\sin\text{x}+10\sec^2\text{x}+8\Big(\frac{\cos^2\text{x}}{\cos\text{x}}\Big)$
$=-4\sin\text{x}+10\sec^2\text{x}+8\cos\text{x}$
Alternate Answer
$(1+2\tan\text{x})(5+4\cos\text{x})=5+4\cos\text{x}+10\tan\text{x}+8\sin\text{x}$
Now we have,
$\frac{\text{d}}{\text{dx}}[(1+2\tan\text{x})(5+4\cos\text{x})]\\=\frac{\text{d}}{\text{dx}}(5+4\cos\text{x}+10\tan\text{x}+8\sin\text{x})$
$=-4\sin\text{x}+10\sec^2\text{x}+8\cos\text{x}$

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