1.
"\\frac{}{}" "\\frac{d ^{}f}{d x^{}} \\sin (x) \\cos (x)=\\cos^2x-\\sin^2x=\\cos2x"
"\\frac{d ^{2}}{d x^{2}} \\sin (x) \\cos (x)=\\frac{d ^{}}{d x^{}}\\cos2x=-2sin2x"
"\\frac{df^3}{dx^3}=\\frac{d}{dx}(-2sin2x)=-4cos2x=-4\\left(\\cos^{2} (x)-\\sin^{2} (x)\\right)"
answer C
2
b or c
"f(x)=\\frac{x^{2}-2x-3}{x+2}" is continous and differentiable on [-1;3]
"f(-1)=0=f(3)"
Hence, according to Rolle's theorem, there's a "c\\in [-1;3]" such that f'(c)=0
Let's find f'(x)
"f'(x)=\\frac{(2x-2)(x+2)-(x^2-2x-3)}{(x+2)^2}=\\frac{2x^2-2x+4x-4-x^2+2x+3}{(x+2)^2}=\\frac{x^2+4x-1}{(x+2)^2}"
"x^2+4x-1=0\\iff x=\\frac{-4\\pm\\sqrt{16+4}}{2}=-2\\pm\\sqrt{5}"
Of these two roots only "c=-2+\\sqrt{5}" lays inside [-1;3]
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