If y = sin x + e^x, then frac{d^2x}{dy^2} = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given $y = \sin x + e^x$. Differentiating with respect to $x$,we get $\frac{dy}{dx} = \cos x + e^x$. Therefore,$\frac{dx}{dy} = \frac{1}{\cos x +

Vedclass · Accepted Answer

$\frac{\sin x - e^x}{(\cos x + e^x)^3}$

Vedclass · Answer

(C) Given $y = \sin x + e^x$. Differentiating with respect to $x$,we get $\frac{dy}{dx} = \cos x + e^x$. Therefore,$\frac{dx}{dy} = \frac{1}{\cos x + e^x} = (\cos x + e^x)^{-1} \dots (i)$. Now,differentiating $\frac{dx}{dy}$ with respect to $y$ using the chain rule: $\frac{d^2x}{dy^2} = \frac{d}{dy} [(\cos x + e^x)^{-1}] = \frac{d}{dx} [(\cos x + e^x)^{-1}] \cdot \frac{dx}{dy}$. $\frac{d^2x}{dy^2} = -1(\cos x + e^x)^{-2} \cdot (-\sin x + e^x) \cdot \frac{dx}{dy}$. Substituting the value of $\frac{dx}{dy}$ from $(i)$: $\frac{d^2x}{dy^2} = -(\cos x + e^x)^{-2} \cdot (-\sin x + e^x) \cdot (\cos x + e^x)^{-1}$. $\frac{d^2x}{dy^2} = \frac{\sin x - e^x}{(\cos x + e^x)^3}$.

If $y = \sin x + e^x,$ then $\frac{d^2x}{dy^2} = $

Similar Questions

If $y=44 x^{45}+45 x^{-44}$,then $y^{\prime \prime}=$

If $y = \frac{\alpha x + \beta}{\gamma x + \delta}$,then $2 y_1 y_3 =$

If $x^2 + y^2 + \sin y = 4$,then the value of $\frac{d^2y}{dx^2}$ at the point $(-2, 0)$ is

If $y=500 e^{7 x}+600 e^{-7 x},$ show that $\frac{d^{2} y}{d x^{2}}=49 y$.

If $y = \log(\log x)$,then $\frac{d^2y}{dx^2}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module