Given that $\displaystyle y=tan2x$ and $\displaystyle \frac{dy}{dx}=2(1+y^2)$, show that $\displaystyle \frac{d^3y}{dx^2}=4(\frac{dy}{dx})^2 +4y\frac{d^2y}{dx^2}$
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Originally Posted by Punch $\displaystyle \frac{dy}{dx}=2(1+y^2)$ Differentiate both sides twice.
Originally Posted by alexmahone Differentiate both sides twice. $\displaystyle \frac{d^2y}{dx^2}=2(2y)\frac{dy}{dx}$ $\displaystyle \frac{d^3y}{dx^2}=4y\frac{d^2y}{dx^2}+\frac{dy}{dx }(4)$ But the answer has a square for dy/dx
Originally Posted by Punch $\displaystyle \frac{d^2y}{dx^2}=2(2y)\frac{dy}{dx}$ $\displaystyle \frac{d^3y}{dx^2}=4y\frac{d^2y}{dx^2}+\frac{dy}{dx }(4)$ But the answer has a square for dy/dx $\displaystyle \frac{d^2y}{dx^2}=4y\frac{dy}{dx}$ $\displaystyle \frac{d^3y}{dx^2}=4\left(y\frac{d^2y}{dx^2}+\frac{ dy}{dx}\frac{dy}{dx}\right)=4(\frac{dy}{dx})^2+4y \frac{d^2y}{dx^2}$
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