Find f'(x) of [sinx².sin²x]/(1 + sinx)
.
ddx([sinx2. sin2x][1+sinx])\frac{d}{dx}\left(\frac{\left[sinx².\:sin²x\right]}{\left[1+sinx\right]}\right)dxd([1+sinx][sinx2.sin2x])
Apply the quotient rule
(fg)′=f′⋅g−g′⋅fg2\left(\frac{f}{g}\right)'=\frac{ f' \cdot g-g'\cdot f}{g^2}(gf)′=g2f′⋅g−g′⋅f
=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}=(1+sin(x))2dxd(sin(x2)sin2(x))(1+sin(x))−dxd(1+sin(x))sin(x2)sin2(x)
=(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}=(1+sin(x))2(cos(x2)⋅2xsin2(x)+sin(2x)sin(x2))(1+sin(x))−cos(x)sin(x2)sin2(x)
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments