Question #167833

Find f'(x) of [sinx².sin²x]/(1 + sinx)


1
Expert's answer
2021-03-02T05:13:08-0500

.

ddx([sinx2.sin2x][1+sinx])\frac{d}{dx}\left(\frac{\left[sinx².\:sin²x\right]}{\left[1+sinx\right]}\right)

Apply the quotient rule

(fg)=fggfg2\left(\frac{f}{g}\right)'=\frac{ f' \cdot g-g'\cdot f}{g^2}


=ddx(sin(x2)sin2(x))(1+sin(x))ddx(1+sin(x))sin(x2)sin2(x)(1+sin(x))2=\frac{\frac{d}{dx}\left(\sin \left(x^2\right)\sin ^2\left(x\right)\right)\left(1+\sin \left(x\right)\right)-\frac{d}{dx}\left(1+\sin \left(x\right)\right)\sin \left(x^2\right)\sin ^2\left(x\right)}{\left(1+\sin \left(x\right)\right)^2}


=(cos(x2)2xsin2(x)+sin(2x)sin(x2))(1+sin(x))cos(x)sin(x2)sin2(x)(1+sin(x))2= \frac{\left(\cos \left(x^2\right)\cdot \:2x\sin ^2\left(x\right)+\sin \left(2x\right)\sin \left(x^2\right)\right)\left(1+\sin \left(x\right)\right)-\cos \left(x\right)\sin \left(x^2\right)\sin ^2\left(x\right)}{\left(1+\sin \left(x\right)\right)^2}



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