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Calculate the derivatives of the functions: 𝑦 = 6𝑥5− 8𝑥4 + 3𝑥3 + 𝑥 − 15

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Answer to a math question Calculate the derivatives of the functions: 𝑦 = 6𝑥5− 8𝑥4 + 3𝑥3 + 𝑥 − 15

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Darrell
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64 Answers
$y '=\frac{ \mathrm{d} }{ \mathrm{d}x} \left( 6{x}^{5}-8{x}^{4}+3{x}^{3}+x-15 \right)$
$y '=\frac{ \mathrm{d} }{ \mathrm{d}x} \left( 6{x}^{5} \right)+\frac{ \mathrm{d} }{ \mathrm{d}x} \left( -8{x}^{4} \right)+\frac{ \mathrm{d} }{ \mathrm{d}x} \left( 3{x}^{3} \right)+\frac{ \mathrm{d} }{ \mathrm{d}x} \left( x \right)-\frac{ \mathrm{d} }{ \mathrm{d}x} \left( 15 \right)$
$y '=6 \times 5{x}^{4}+\frac{ \mathrm{d} }{ \mathrm{d}x} \left( -8{x}^{4} \right)+\frac{ \mathrm{d} }{ \mathrm{d}x} \left( 3{x}^{3} \right)+\frac{ \mathrm{d} }{ \mathrm{d}x} \left( x \right)-\frac{ \mathrm{d} }{ \mathrm{d}x} \left( 15 \right)$
$y '=6 \times 5{x}^{4}-8 \times 4{x}^{3}+\frac{ \mathrm{d} }{ \mathrm{d}x} \left( 3{x}^{3} \right)+\frac{ \mathrm{d} }{ \mathrm{d}x} \left( x \right)-\frac{ \mathrm{d} }{ \mathrm{d}x} \left( 15 \right)$
$y '=6 \times 5{x}^{4}-8 \times 4{x}^{3}+3 \times 3{x}^{2}+\frac{ \mathrm{d} }{ \mathrm{d}x} \left( x \right)-\frac{ \mathrm{d} }{ \mathrm{d}x} \left( 15 \right)$
$y '=6 \times 5{x}^{4}-8 \times 4{x}^{3}+3 \times 3{x}^{2}+1-\frac{ \mathrm{d} }{ \mathrm{d}x} \left( 15 \right)$
$y '=6 \times 5{x}^{4}-8 \times 4{x}^{3}+3 \times 3{x}^{2}+1-0$
$y '=30{x}^{4}-32{x}^{3}+9{x}^{2}+1$

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