Math, asked by Jinsonjose, 1 year ago

second derivative of log(1+sinx)

Answers

Answered by MarkAsBrainliest
15

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}}

#MarkAsBrainliest

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