Question

second derivative of log(1+sinx)

Answer 1

Answer :

Let, y = log (1 + sinx)

Now, differentiating both sides with respect to x, we get

$\bold{\frac{dy}{dx} = \frac{d}{dx} \{log(1 + sinx) \}}$

$\bold{= \frac{1}{1+sinx} * \frac{d}{dx}(1+sinx)}$

$\bold{= \frac{1}{1+sinx} * cosx}$

$\bold{= \frac{cosx}{1+sinx}}$

$\bold{= \frac{cosx}{1+sinx} * \frac{1-sinx}{1-sinx}}$

$\bold{= \frac{cosx(1-sinx)}{1-sin^{2}x}}$

$\bold{= \frac{cosx (1-sinx)}{cos^{2}x}}$

$\bold{=\frac{1-sinx}{cosx}}$

$\bold{= \frac{1}{cosx} - \frac{sinx}{cosx}}$

$\bold{= secx - tanx}$

$\to \boxed{\bold{\frac{d}{dx} \{log(1 + sinx) \}=secx-tanx}}$

Again, differentiating both sides with respect to x, we get

$\bold{\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}(secx-tanx)}$

$\bold{= \frac{d}{dx}(secx) - \frac{d}{dx}(tanx)}$

$\bold{= secx\:tanx - sec^{2}x}$

$\to \boxed{\bold{\frac{d^2}{dx^{2}}\{log(1+sinx)\}=secx\:tanx - sec^{2}x}}$

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