If-y-2sinx-tanx-prove-that-d-2-y-dx-2-2sinx-sec-3-x-1-Any-detailed-solution-please-

Question Number 141769 by 7770 last updated on 23/May/21

If y = 2 sinx + tanx, prove that

\frac{d^{2} y}{{dx}^{2}} = 2 sinx (\sec^{3} x - 1)

Any detailed solution please .

Answered by physicstutes last updated on 23/May/21

y = 2 \sin x + \tan x

\frac{d y}{d x} = 2 \cos x + \sec^{2} x

\frac{d^{2} y}{{d x}^{2}} = - 2 \sin x + \frac{(\cos^{2} x) (0) - (1) (2 \cos x) (- \sin x)}{\cos^{4} x}

\frac{d^{2} y}{{d x}^{2}} = \frac{2 \sin x \cos x}{\cos^{4} x} - 2 \sin x = \frac{2 \sin x}{\cos^{3} x} - 2 \sin x = 2 \sin x (\frac{1}{\cos^{3} x} - 1) = 2 \sin x (\sec^{3} x - 1)

as required .