Let g(x) neq 0, g^{prime}(x) neq 0, f(x) neq | vedclass

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(D) Given $F(x)=f(x) g(x)$.By the product rule, $F^{\prime}(x)=f^{\prime}(x) g(x)+f(x) g^{\prime}(x)$.We are given $G(x)=f^{\prime}(x) g^{\prime}(x)$.

Vedclass · Accepted Answer

$\frac{f^{\prime}}{f}+\frac{g}{g^{\prime}}+\frac{f}{f^{\prime}}+\frac{g^{\prime}}{g}$

Vedclass · Answer

(D) Given $F(x)=f(x) g(x)$.By the product rule, $F^{\prime}(x)=f^{\prime}(x) g(x)+f(x) g^{\prime}(x)$.We are given $G(x)=f^{\prime}(x) g^{\prime}(x)$.Since $F^{\prime}(x)=G(x) H(x)$, we have $H(x)=\frac{F^{\prime}(x)}{G(x)}=\frac{f^{\prime}(x) g(x)+f(x) g^{\prime}(x)}{f^{\prime}(x) g^{\prime}(x)}=\frac{g(x)}{g^{\prime}(x)}+\frac{f(x)}{f^{\prime}(x)}$.Also, $F^{\prime}(x)=F(x) K(x)$, so $K(x)=\frac{F^{\prime}(x)}{F(x)}=\frac{f^{\prime}(x) g(x)+f(x) g^{\prime}(x)}{f(x) g(x)}=\frac{f^{\prime}(x)}{f(x)}+\frac{g^{\prime}(x)}{g(x)}$.Therefore, $H(x)+K(x)=\frac{f(x)}{f^{\prime}(x)}+\frac{g(x)}{g^{\prime}(x)}+\frac{f^{\prime}(x)}{f(x)}+\frac{g^{\prime}(x)}{g(x)}$.

Let $g(x) \neq 0, g^{\prime}(x) \neq 0, f(x) \neq 0, f^{\prime}(x) \neq 0$. If $F(x)=f(x) g(x)$, $G(x)=f^{\prime}(x) g^{\prime}(x)$, $F^{\prime}(x)=G(x) H(x)$, and $F^{\prime}(x)=F(x) K(x)$, then $H(x)+K(x)=$

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