Question

solve Q. 25, 26, 27, 28. For 10 points
LEVEL- CLASS 11 IIT PREPARATION
DIFFERENTIATUON​

Answer 1

Answer:

27. option (a) is correct

28. option (a) is correct

Step-by-step explanation:

27.

$y=a\:e^{mx}+b\:e^{-mx}$

Differentiate with respect to 'x'

$\frac{dy}{dx}=a\:e^{mx}(m)+b\:e^{-mx}(-m)$

Differentiate once again with respect to 'x'

$\frac{d^2y}{dx^2}=a\:e^{mx}(m^2)+b\:e^{-mx}(-m)^2$

$\frac{d^2y}{dx^2}=a\:e^{mx}(m^2)+b\:e^{-mx}(m^2)$

$\frac{d^2y}{dx^2}=m^2[a\:e^{mx}+b\:e^{-mx}]$

$\frac{d^2y}{dx^2}=m^2y$

option (a) is correct

28.

$y=log[x+\sqrt{1+x^2}]$

Differentiate with respect to 'x'

$\frac{dy}{dx}=\frac{1}{x+\sqrt{1+x^2}}*(1+\frac{2x}{2\sqrt{1+x^2}})$

$\frac{dy}{dx}=\frac{1}{x+\sqrt{1+x^2}}*(1+\frac{x}{\sqrt{1+x^2}})$

$\frac{dy}{dx}=\frac{1}{x+\sqrt{1+x^2}}*(\frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}})$

$\frac{dy}{dx}=\frac{1}{\sqrt{1+x^2}}$

$\frac{dy}{dx}=(1+x^2)^{\frac{-1}{2}}$

Differentiate once again with respect to x

$\frac{d^2y}{dx^2}=\frac{-1}{2}(1+x^2)^{\frac{-3}{2}}(2x)$

$\frac{d^2y}{dx^2}=-(1+x^2)^{\frac{-3}{2}}(x)$

$\frac{d^2y}{dx^2}=-x\frac{1}{(1+x^2)^{\frac{3}{2}}}$

$\frac{d^2y}{dx^2}=-x\frac{1}{{1+x^2}\sqrt{1+x^2}}$

$(1+x^2)\frac{d^2y}{dx^2}=-x\frac{1}{\sqrt{1+x^2}}$

$(1+x^2)\frac{d^2y}{dx^2}=-x\frac{dy}{dx}$

$(1+x^2)\frac{d^2y}{dx^2}+x\frac{dy}{dx}=0$

option (a) is correct