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If $y=x+\tan x$, then $\cos ^2 x \frac{d^2 y}{d x^2}+2 x$ is equal to
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Verified Answer
The correct answer is:
2y
Given, $y=x+\tan x$
Differentiating, $\frac{d y}{d x}=1+\sec ^2 x$
Again differentiating, $\frac{d^2 y}{d x^2}=0+2 \sec x \sec x \tan x$
$$
=2 \sec ^2 x \cdot \tan x
$$
Now, $\cos ^2 x \cdot \frac{d^2 y}{d x^2}+2 x=\cos ^2 x \cdot 2 \sec ^2 x \cdot \tan x+2 x$
$$
=2(x+\tan x)=2 y
$$
Differentiating, $\frac{d y}{d x}=1+\sec ^2 x$
Again differentiating, $\frac{d^2 y}{d x^2}=0+2 \sec x \sec x \tan x$
$$
=2 \sec ^2 x \cdot \tan x
$$
Now, $\cos ^2 x \cdot \frac{d^2 y}{d x^2}+2 x=\cos ^2 x \cdot 2 \sec ^2 x \cdot \tan x+2 x$
$$
=2(x+\tan x)=2 y
$$
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