Solution: Provided: $e^{y}(x+1)=1$ Therefore, $e^{y}=\frac{1}{x+1}$ On both the sides taking log , we get $\log e^{y}=\log \frac{1}{x+1}$ $y \log e=\log 1-\log (x+1)$ $y=-\log (x+1)$ $\frac{d y}{d...

Solution: $y=500 e^{7 x}+600 e^{-7 x} \ldots \ldots \ldots . .(1)$ $\frac{d y}{d x}=500 e^{7 x}(7)+600 e^{-7 x}(-7)$ $=500(7) e^{7 x}-600(7) e^{7 x}$ So now, $\frac{d^{2} y}{d x^{2}}=500(7) e^{7...

Solution: We need to Prove: $\frac{d^{2} y}{d x^{2}}-(m+n) \frac{d y}{d x}+m n y=0$ $\frac{d y}{d x}=\mathrm{A} e^{m x} \frac{d}{d x}(m x)+\mathrm{B} e^{n x} \frac{d}{d x}(n x)\left[\because...

Solution: The provided function is $y=3 \cos (\log x)+4 \sin (\log x) \ldots .(1)$ With respect to $x$ derivate , we obtain $\frac{d y}{d x}=y_{1}=-3 \sin (\log x) \frac{d}{d x} \log x+4 \cos (\log...

Solution: Provided: $y=\cos ^{-1} x$ or $x=\cos v$ or $x=\cos y \ldots \ldots \ldots . .(1)$ $\frac{d y}{d x}=\frac{-1}{\sqrt{1-x^{2}}}$[From (1)] $=\frac{-1}{\sqrt{\sin ^{2} y}}=\frac{-1}{\sin...

Solution: Let's take $y=5 \cos x-3 \sin x \ldots$...(1) $\frac{d y}{d x}=-5 \sin x-3 \cos x$ So now, $\frac{d^{2} y}{d x^{2}}=-5 \cos x+3 \sin x$ $={-(5 \cos x-3 \sin x)=-y}$ [From (1)] $\frac{d^{2}...

Solution: Le's take $y=\sin (\log x)$ $\frac{d y}{d x}=\cos (\log x) \frac{d}{d x}(\log x)$ $=\cos (\log x) \cdot \frac{1}{x}$ $=\frac{\cos (\log x)}{x}$ So now, $\frac{d^{2} y}{d x^{2}}=\frac{x...

Solution: Let's take ${ }^{y=\log (\log x)}$ $\frac{d y}{d x}=\frac{1}{\log x} \frac{d}{d x} \log x$ $\left[\because \frac{d}{d x} \log f(x)=\frac{1}{f(x)} \frac{d}{d x} f(x)\right]$ $=\frac{1}{\log...

Solution: Let's take $y=\tan ^{-1} x$ $\frac{d y}{d x}=\frac{1}{1+x^{2}}$ $\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(\frac{1}{1+x^{2}}\right)$ $=\frac{\left(1+x^{2}\right) \frac{d}{d x}(1)-1...

Solution: Let's take $y=e^{6 x} \cos 3 x$ $\frac{d y}{d x}=e^{6 x} \frac{d}{d x} \cos 3 x+\cos 3 x \frac{d}{d x} e^{6 x}$ $=e^{6 x}(-\sin 3 x) \frac{d}{d x}(3 x)+\cos 3 x \cdot e^{6 x} \frac{d}{d...

Solution: Let's take $y=e^{x} \sin 5 x$ $\frac{d y}{d x}=e^{x} \frac{d}{d x} \sin 5 x+\sin 5 x \frac{d}{d x} e^{x}$ $=e^{x} \cos 5 x \frac{d}{d x} 5 x+\sin 5 x \cdot e^{x}$ $=e^{x} \cos 5 x \times...

Solution: Let's take $y=x^{3} \log x$ $\frac{d y}{d x}=x^{3} \frac{d}{d x} \log x+\log x \frac{d}{d x} x^{3}$ $=x^{3} \cdot \frac{1}{x}+\log x\left(3 x^{2}\right)$ $=x^{2}+3 x^{2} \log x$ So now,...

Solution: Let's take $y=\log x$ $\frac{d y}{d x}=\frac{1}{x}$ $\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(\frac{1}{x}\right)=\frac{d}{d x} x^{-1}$ $\frac{d^{2} y}{d x^{2}}=(-1)...

Solution: Let's take $y=x \cos x$ $\frac{d y}{d x}=x \frac{d}{d x} \cos x+\cos x \frac{d}{d x} x$ $=-x \sin x+\cos x$ So now, $\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(\frac{d y}{d...

Solution: Let's take $y=x^{20}$ With respect to $x$ derivate y , we have $\frac{d y}{d x}=20 x^{19}$ With respect to $x$ derivate dy/dx , we obtain $\frac{d^{2} y}{d x^{2}}=\frac{d}{d...

Solution: Let's take $y=x^{2}+3 x+2$ Now the first derivative: $\frac{d y}{d x}=2 x+3 \times 1+0=2 x+3$ The second derivative: $\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(\frac{d y}{d x}\right)=2...