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vpaliwal977
@vpaliwal977
July 2023
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7. Find the quadratic polynomial in each case, with the given numbers as its zeroes (i) 3,-5 (ii) 5,-5 (iv) 3,4 (iii) -9, L 9
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mathlovers1729
Answer:
[tex] y = {e}^{4x} + {x}^{2} \\ \\ \implies \: \frac{dy}{dx} = \frac{d}{dx} {e}^{4x} \frac{d}{dx} 4x + \frac{d}{dx} {x}^{2} \\ \\ \implies \: \frac{dy}{dx} = {e}^{4x} .4 + 2x \\ \\ \implies \: \frac{dy}{dx} = 4{e}^{4x} + 2x \\ [/tex]
derivative =4e^4x+2x
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Answer:
[tex] y = {e}^{4x} + {x}^{2} \\ \\ \implies \: \frac{dy}{dx} = \frac{d}{dx} {e}^{4x} \frac{d}{dx} 4x + \frac{d}{dx} {x}^{2} \\ \\ \implies \: \frac{dy}{dx} = {e}^{4x} .4 + 2x \\ \\ \implies \: \frac{dy}{dx} = 4{e}^{4x} + 2x \\ [/tex]
derivative =4e^4x+2x