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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(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 $\lim _{x \rightarrow-1}(\frac{d y}{d x})=$

Similar Questions

The derivative of $F[f\{ \phi (x)\} ]$ with respect to $x$ is:

$\frac{d}{dx}[e^{ax} \cos(bx + c)] = ?$

$\frac{d}{d x}\left[\cos ^2\left(\cot ^{-1} \sqrt{\frac{2+x}{2-x}}\right)\right]$ is

The derivative of $y = (1 - x)(2 - x)...(n - x)$ at $x = 1$ is equal to

If $y = f(x^2 + 2)$ and $f'(3) = 5$,then $\frac{dy}{dx}$ at $x = 1$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module