Question

If $e^{g\left(y\right)}-e^{-g\left(y\right)}=2f\left(x\right)$ then $\frac{dy}{dx}=$

• A. (1) $\frac{f^1\left(x\right)}{g^1\left(y\right)}\sqrt{1+f{\left(x\right)}^2}$
• B. (2) $\frac{f^1\left(x\right)}{g^1\left(y\right)}\frac{1}{\sqrt{1+{\left(f\left(x\right)\right)}^2}}$
• C. (3)$\frac{f^1\left(x\right)}{g\left(y\right)}\frac{1}{\sqrt{1+{\left(f\left(x\right)\right)}^2}}$
• D. (4) $\frac{f\left(x\right)}{g^1\left(y\right)}\sqrt{1+f{\left(x\right)}^2}$

Answer 1

The correct option is B (2) $\frac{f^1\left(x\right)}{g^1\left(y\right)}\frac{1}{\sqrt{1+{\left(f\left(x\right)\right)}^2}}$

$e^{g\left(y\right)}-$$e^{-g\left(y\right)}=2f\left(x\right)$

Differentiating on both sides

$e^{g\left(y\right)}{g}^1\left(y\right)\frac{dy}{dx}$$+e^{-g\left(y\right)}{g}^1\left(y\right)\frac{dy}{dx}=2f^1\left(x\right)$

$\frac{dy}{dx}[e^{g\left(y\right)}+e^{-g\left(y\right)}]=\frac{2f^1\left(x\right)}{g^1\left(y\right)}$

$\frac{dy}{dx}=\frac{2f^1\left(x\right)}{g^1\left(y\right)}\frac{1}{e^{g\left(y\right)}+e^{-g\left(y\right)}}$

$f\left(x\right)=$$\frac{e^{g\left(y\right)}+e^{-g\left(y\right)}}{2}$

$f(x{)}^2=$$\frac{e^{2g\left(y\right)}+e^{-2g\left(y\right)}-2}{4}$

$1+f(x{)}^2=$$1+\frac{e^{2g\left(y\right)}+e^{-2g\left(y\right)}-2}{4}$

$1+f(x{)}^2=$$\frac{e^{2g\left(y\right)}+e^{-2g\left(y\right)}-2+4}{4}$

$1+f(x{)}^2=$$\frac{(e^{g\left(y\right)}+e^{-g\left(y\right)}{)}^2}{2^2}$

$\sqrt{1+f(x{)}^2}=$$\frac{e^{g\left(y\right)}+e^{-g\left(y\right)}}{2}$

$2\sqrt{1+f(x{)}^2}=$$e^{g\left(y\right)}+e^{-g\left(y\right)}$

Now we have,

$\frac{dy}{dx}=\frac{2f^1\left(x\right)}{g^1\left(y\right)}\frac{1}{e^{g\left(y\right)}+e^{-g\left(y\right)}}$

$\frac{dy}{dx}=\frac{2f^1\left(x\right)}{g^1\left(y\right)}\frac{1}{2\sqrt{1+f(x{)}^2}}$

$\frac{dy}{dx}=\frac{f^1\left(x\right)}{g^1\left(y\right)}\frac{1}{\sqrt{1+f(x{)}^2}}$