Find ππ¦ ππ₯ if π¦ = π’ 3 β 3π’ 2 + 1 and π’ = π₯ 2 + 2Β
"\\frac{dy}{dx}=\\frac{dy}{du}\\frac{du}{dx}=\\left( 3u^2-6u \\right) \\cdot 2x=6x\\left( \\left( x^2+2 \\right) ^2-2\\left( x^2+2 \\right) \\right) =\\\\=6x\\left( x^4+2x^2 \\right)"
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