If y=(x-1)(x+2)(x^2+5)(x^4+8),then lim _{x ri | vedclass

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(B) Given $y = (x-1)(x+2)(x^2+5)(x^4+8)$.Using the product rule for differentiation $\frac{d}{dx}(uvwz) = u'vwz + uv'wz + uvw'z + uvwz'$,we get:$\frac

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$30$

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(B) Given $y = (x-1)(x+2)(x^2+5)(x^4+8)$.Using the product rule for differentiation $\frac{d}{dx}(uvwz) = u'vwz + uv'wz + uvw'z + uvwz'$,we get:$\frac{dy}{dx} = (1)(x+2)(x^2+5)(x^4+8) + (x-1)(1)(x^2+5)(x^4+8) + (x-1)(x+2)(2x)(x^4+8) + (x-1)(x+2)(x^2+5)(4x^3)$.Now,evaluate the limit as $x \rightarrow -1$:For $x = -1$:Term $1$: $(1)(-1+2)((-1)^2+5)((-1)^4+8) = (1)(1)(6)(9) = 54$.Term $2$: $(-1-1)(1)((-1)^2+5)((-1)^4+8) = (-2)(1)(6)(9) = -108$.Term $3$: $(-1-1)(-1+2)(2(-1))((-1)^4+8) = (-2)(1)(-2)(9) = 36$.Term $4$: $(-1-1)(-1+2)((-1)^2+5)(4(-1)^3) = (-2)(1)(6)(-4) = 48$.Summing these values: $54 - 108 + 36 + 48 = 30$.

If $y=(x-1)(x+2)(x^2+5)(x^4+8)$,then $\lim _{x \rightarrow-1}(\frac{d y}{d x})=$

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