If $x^{y}=\mathrm{e}^{x-y}$ then $\frac{d y}{d x}=\ldots \ldots$
यदि $x^{y}=\mathrm{e}^{x-y}$ तो $\frac{d y}{d x}=\ldots \ldots$
If $x^{\mathrm{a}} \cdot \mathrm{y}^{\mathrm{b}}=(x+\mathrm{y})^{\mathrm{a}+\mathrm{b}} \cdot$ then $\frac{d y}{d x}$ equals
यदि $x^{\mathrm{a}} \cdot \mathrm{y}^{\mathrm{b}}=(x+\mathrm{y})^{\mathrm{a}+\mathrm{b}} \cdot$ तो $\frac{d y}{d x}$ =
If $\mathrm{y}=e^{x+e^{x+\ldots+ up t o \infty}}$ then $\frac{d y}{d x}=\ldots . .$
यदि $\mathrm{y}=e^{x+e^{x+\ldots+t o \infty}}$ तो $\frac{d y}{d x}=\ldots . .$
If $x^{y}=y^{x}$ and $\frac{d y}{d x}=f(x, y)\left(\frac{x \log y-y}{y \log x-x}\right)$ then $f(x, y)=\ldots \ldots$
यदि $x^{y}=y^{x}$ and $\frac{d y}{d x}=f(x, y)\left(\frac{x \log y-y}{y \log x-x}\right)$ तो $f(x, y)=\ldots \ldots$
If $y=\cos ^{-1} \sqrt{\frac{\sqrt{1+x^{2}}+1}{2 \sqrt{1+x^{2}}}}$ then $\frac{d y}{d x}=\ldots \ldots$
यदि $y=\cos ^{-1} \sqrt{\frac{\sqrt{1+x^{2}}+1}{2 \sqrt{1+x^{2}}}}$ तो $\frac{d y}{d x}=\ldots \ldots$
If $y=\cos ^{-1}\left(\frac{x-x^{-1}}{x+x^{-1}}\right)$ then $\frac{d y}{d x}$ will be:
यदि $y=\cos ^{-1}\left(\frac{x-x^{-1}}{x+x^{-1}}\right)$ तो $\frac{d y}{d x}$ =
If $y=x \log y$ then which of the following is correct?
यदि $y=x \log y$ तो निम्नलिखित में से कौन सा सही है?
If $y=\sqrt{x}+\frac{1}{\sqrt{x}}$ then $2 x \frac{d y}{d x}+y$ is equal to
यदि $y=\sqrt{x}+\frac{1}{\sqrt{x}}$ तो $2 x \frac{d y}{d x}+y$ = . . . . . . . .
If $y=e^{\tan x}$ then the value of $\cos ^{2} x \frac{d^{2} y}{d x^{2}}-$ $(1+\sin 2 x) \frac{d y}{d x}=\ldots . .$
यदि $y=e^{\tan x}$ तब $\cos ^{2} x \frac{d^{2} y}{d x^{2}}-$ $(1+\sin 2 x) \frac{d y}{d x}=\ldots . .$
If $y=e^{\operatorname{ax}} \sin b x$ then $\frac{d^{2} y}{d x^{2}}-2 x \frac{d y}{d x}+k g=0$ where $\mathrm{k}$ equals:
यदि $y=e^{\operatorname{ax}} \sin bx$ तब $\frac{d^{2} y}{dx^{2}}-2 x \frac{dy}{dx}+kg= 0$ जहां $\mathrm{k}$ का मान क्या होगा ?
Enter Your Mobile No. To Login/Register
⇐ Go Back to change the mobile no.
Didn't receive OTP? Resend OTP -OR- Voice call Call Again/Resend OTP in 30 Seconds
