Find second derivative of ex = tan2y with respect to x.

1.	Find second derivative of ex = tan2y with respect to x.
Answer» e^x = tan 2y----(1) differentiate (1) w.r.t. x we get e^x = 2 sec²2y.\(\frac{dy}{dx}\)----(2) differentiate (2) w.r.t. x, we get e^x = 2sec²2y \(\frac{d^2y}{dx^2}+\) 8sec 2y. sec 2y tan 2y \((\frac{dy}{dx})^2\) = 2 sec²2y.\(\frac{d^2y}{dx^2}+\) 8 sec²2y tan 2y \((\frac{e^x}{2sec^22y})^2\) (From (2)) = 2 sec²2y \(\frac{d^2y}{dx^2}+\) 2 \(\frac{tan2y}{sec^22y}e^{2x}\) ⇒ 2 sec²2y.\(\frac{d^2y}{dx^2}\) = e^x - \(\frac{2tan2y}{sec^22y}e^{2x}\) ⇒ 2(1 + tan²2y)\(\frac{d^2y}{dx^2}\) = e^x - \(\frac{2.e^x.e^{2x}}{1+tan^22y}\) (\(\because\) 1 + tan²\(\theta\) = sec²\(\theta\) and From (1)) ⇒ 2(1 + e^2x) \(\frac{d^2y}{dx^2}\) = e^x - \(\frac{2e^{3x}}{1+e^{2x}}\) ⇒ \(\frac{d^2y}{dx^2}\) = \(\frac{e^{3x}+e^x-2e^{3x}}{2(1+e^{2x})^2}\)

Answer»

e^x = tan 2y----(1)

differentiate (1) w.r.t. x we get

e^x = 2 sec²2y.\(\frac{dy}{dx}\)----(2)

differentiate (2) w.r.t. x, we get

e^x = 2sec²2y \(\frac{d^2y}{dx^2}+\) 8sec 2y. sec 2y tan 2y \((\frac{dy}{dx})^2\)

= 2 sec²2y.\(\frac{d^2y}{dx^2}+\) 8 sec²2y tan 2y \((\frac{e^x}{2sec^22y})^2\) (From (2))

= 2 sec²2y \(\frac{d^2y}{dx^2}+\) 2 \(\frac{tan2y}{sec^22y}e^{2x}\)

⇒ 2 sec²2y.\(\frac{d^2y}{dx^2}\) = e^x - \(\frac{2tan2y}{sec^22y}e^{2x}\)

⇒ 2(1 + tan²2y)\(\frac{d^2y}{dx^2}\) = e^x - \(\frac{2.e^x.e^{2x}}{1+tan^22y}\) (\(\because\) 1 + tan²\(\theta\) = sec²\(\theta\) and From (1))

⇒ 2(1 + e^2x) \(\frac{d^2y}{dx^2}\) = e^x - \(\frac{2e^{3x}}{1+e^{2x}}\)

⇒ \(\frac{d^2y}{dx^2}\) = \(\frac{e^{3x}+e^x-2e^{3x}}{2(1+e^{2x})^2}\)

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