Question

double differentiate y=secx -tanx prove that (cos)d2y/dx2=y²​

Answer 1

Step-by-step explanation:

$\huge\bf{\mid{\overline{\underline{ANSWER}\mid}}}$

$y ={ \sec}(x) - \tan(x) \\ \\ y = \frac{1}{\cos(x)} - \frac{\sin(x)}{\cos(x)} \\ \\ \boxed{ y = \frac{1 - \sin(x)}{cos(x)}}.......(1) \\ \\ \frac{dy}{dx} = \frac{\cos(x) (- \cos(x)) - (1 - \sin(x)) \times (- \sin(x))}{{\cos}^{2}(x)} \\ \\ When \: \: Simplified \: \: further...\\ \\ \frac{dy}{dx} = \frac{ - {\cos}^{2}(x) + \sin(x) - {\sin}^{2}(x) }{{\cos}^{2}(x)} \\ \\ \frac{dy}{dx} = \frac{\sinx - 1}{{\cos}^{2}(x)} \\ \\ \frac{{d}^{2}y}{d{x}^{2}} = \frac{{\cos}^{2}(x) \times(\cos(x)) - ( {\sinx} - 1) (2 \cos(x) \times (- \sin(x))}{{\cos}^{4}(x)} \\ \\ When \: \: Simplified \: \: further... \\ \\ \frac{{d}^{2}y}{d{x}^{2}} = \frac{ 1 + {\sin}^{2}(x) - \sin(x)}{{\cos}^{3}(x)} \\ \\ \cos(x) \frac{{d}^{2}y}{d{x}^{2}} = \frac{1 + {\sin}^{2}(x) - 2 sin(x) }{{\cos}^{2}(x)} \\ \\ \cos(x) \frac{{d}^{2}y}{d{x}^{2}} = \frac{{1 - \sin(x)}^{2}}{{\cos}^{2}(x)}$$\cos(x) \frac{{d}^{2}y}{d{x}^{2}} = \frac{{1 - \sin(x)}^{2}}{{\cos(x)}{2}} \\ \\ cos(x) \frac{{d}^{2}y}{d{x}^{2}} = {y}^{2}$