How To Answer this Calculus Question and Find dy dx:?

4.
dy/dx
= d(x³eˣ)/dy
= x³(deˣ/dx) + eˣ(dx³/dx)
= x³(eˣ) + eˣ(3x²)
= x³eˣ + 3x²eˣ

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5.
dy/dx
= d[(cosx)(eˣ)]/dx
= (cosx)(deˣ/dx) + (eˣ)(dcosx/dx)
= (cosx)(eˣ) + (eˣ)(-sinx)
= (cosx)(eˣ) - (sinx)(eˣ)

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6.
dy/dx
= (d/dx)[(x³ + 1)(√x)/x²]
= {x²[(x³ + 1)(√x)]' - [(x³ + 1)(√x)](x²)'}/(x²)²
= {x²[(x³ + 1)(√x)' + (√x)(x³ + 1)'] - [(x³ + 1)(√x)](2x)}/x⁴
= {x²[(x³ + 1)(1/2√x) + (√x)(3x²)] - 2x(x³ + 1)√x]}/x⁴
= [(x√x)(x³ + 1)/2 + 3x⁴√x - 2x(x³ + 1)√x]/x⁴
= (√x)(x⁴ + x + 6x⁴ - 4x⁴ - 4x)/(2x⁴)
= (√x)(3x⁴ - 3x)/(2x⁴)
= 3(√x)(x³ - 1)/(2x³)

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7.
f'(x)
= [g(x)h(x)]'
= g(x)h'(x) + g'(x)h(x)

f'(10)
= g(10)h'(10) + g'(10)h(10)
= (-4)×(35) + (0)(560)
= -140

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8.
g'(x)
= (x + √x)(eˣ)
= (x + √x)(eˣ)' + (x +√x)'(eˣ)
= (x + √x)(eˣ) + {1 + [1/(2√x)]}(eˣ)
= (2x² + 2x√x)(eˣ)/(2x) + [2x + √x](eˣ)/(2x)
= (eˣ)(2x² + 2x√x + 2x + √x)/(2x)