Question

d/dx[(x^2+1)^sinx] .

Answer 1

Solution:

To find $\frac{d}{dx} [(x^2+1)^{\sin x }]$

Let $y=(x^2+1)^{\sin x } ...(i)$

Taking log to both sides,

$\log y=\log (x^2+1)^{\sin x } \\ \log y=\sin x \log (x^2+1) [\because \log m^n=n \log m]$

Now differentiating both sides w.r.t $x$ using product rule on right side,

$\frac d {dx} \log y=\sin x \frac d {dx} [\log (x^2+1)]+\log (x^2+1) \frac d {dx}\sin x$

$\frac 1 y\frac {dy} {dx}=\sin x[\frac 1 {x^2+1} \frac d {dx} (x^2+1)]+\log (x^2+1) \cos x \\ \frac 1 y\frac {dy} {dx}=\sin x[\frac 1 {x^2+1}\times 2x]+\log (x^2+1) \cos x \\ \frac {dy} {dx}=y[\frac {2x\sin x} {x^2+1}+\log (x^2+1) \cos x]$

Now using (i),

$\frac {dy}{dx}=(x^2+1)^{\sin x}[\frac {2x\sin x} {x^2+1}+\log (x^2+1) \cos x]$

Answer:

$(x^2+1)^{\sin x}[\frac {2x\sin x} {x^2+1}+\log (x^2+1) \cos x]$