Differentiation of (x^2-5x+8) times (x^3+7x+9 | vedclass

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(D) The given function is a product of two polynomials,$f(x) = (x^2-5x+8)(x^3+7x+9)$.$1$. Product Rule: We can use the product rule $\frac{d}{dx}[u(x)

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All of the options are correct

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(D) The given function is a product of two polynomials,$f(x) = (x^2-5x+8)(x^3+7x+9)$.$1$. Product Rule: We can use the product rule $\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$,where $u(x) = x^2-5x+8$ and $v(x) = x^3+7x+9$.$2$. Expansion: We can expand the product to obtain a single polynomial of degree $5$ and then differentiate each term using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$.$3$. Logarithmic Differentiation: Since the function is a product of factors,we can take the natural logarithm on both sides,$\ln(y) = \ln(x^2-5x+8) + \ln(x^3+7x+9)$,and then differentiate implicitly.Since all three methods are valid and applicable,the correct answer is $D$.

Column $I$	Column $II$
$(A)$ $f(x) = x\|x\|$	$(p)$ continuous in $(-1, 1)$
$(B)$ $f(x) = \sqrt{\|x\|}$	$(q)$ differentiable in $(-1, 1)$
$(C)$ $f(x) = x + [x]$	$(r)$ strictly increasing in $(-1, 1)$
$(D)$ $f(x) = \|x - 1\| + \|x + 1\|$	$(s)$ not differentiable at least at one point in $(-1, 1)$

Differentiation of $(x^2-5x+8) \times (x^3+7x+9)$ can be done by

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