Derivatives — MATHS STD 11 Science — Question

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,

$\frac{\text{d}}{\text{dx}}[(1+2\tan\text{x})(5+4\cos\text{x})]\\=\frac{\text{d}}{\text{dx}}(5+4\cos\text{x}+10\tan\text{x}+8\sin\text{x})$

$=-4\sin\text{x}+10\sec^2\text{x}+8\cos\text{x}$