let f(x)=(6x - 7x^3)(7 + √x). find f(x). f(x)=

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u(x)v(x)$, then $y'=u'(x)v(x)+u(x)v'(x)$. Let $u(x)=6x - 7x^{3}$ and $v(x)=7+\sqrt{x}=7 + x^{\frac{1}{2}}$.

Step2: Find $u'(x)$

Differentiate $u(x)=6x - 7x^{3}$ with respect to $x$. Using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, we have $u'(x)=\frac{d}{dx}(6x)-\frac{d}{dx}(7x^{3})=6-21x^{2}$.

Step3: Find $v'(x)$

Differentiate $v(x)=7 + x^{\frac{1}{2}}$ with respect to $x$. $\frac{d}{dx}(7)=0$ and $\frac{d}{dx}(x^{\frac{1}{2}})=\frac{1}{2}x^{-\frac{1}{2}}$. So $v'(x)=\frac{1}{2\sqrt{x}}$.

Step4: Calculate $f'(x)$

$f'(x)=u'(x)v(x)+u(x)v'(x)=(6 - 21x^{2})(7+\sqrt{x})+(6x - 7x^{3})\frac{1}{2\sqrt{x}}$. Expand the first term: $(6 - 21x^{2})(7+\sqrt{x})=42 + 6\sqrt{x}-147x^{2}-21x^{\frac{5}{2}}$. Expand the second term: $(6x - 7x^{3})\frac{1}{2\sqrt{x}}=\frac{6x}{2\sqrt{x}}-\frac{7x^{3}}{2\sqrt{x}} = 3\sqrt{x}-\frac{7}{2}x^{\frac{5}{2}}$. Combine like terms: $f'(x)=42+(6\sqrt{x}+3\sqrt{x})-147x^{2}+(-21x^{\frac{5}{2}}-\frac{7}{2}x^{\frac{5}{2}})=42 + 9\sqrt{x}-147x^{2}-\frac{42 + 7}{2}x^{\frac{5}{2}}=42 + 9\sqrt{x}-147x^{2}-\frac{49}{2}x^{\frac{5}{2}}$.

Answer:

$42 + 9\sqrt{x}-147x^{2}-\frac{49}{2}x^{\frac{5}{2}}$