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g^(')(x)=(2x+5y)/(x)
Is
g(x)=x^(5)-2x a solution to the above equation?
Choose 1 answer:
(A) Yes
(B)
No

$g^{\prime}(x)=\frac{2 x+5 y}{x}$ $\newline$ Is $g(x)=x^{5}-2 x$ a solution to the above equation? $\newline$ Choose $1$ answer: $\newline$ (A) Yes $\newline$ (B) $\mathrm{No}$

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Is (x, y) a solution to the nonlinear equation?

Full solution

Q.

g^{\prime}(x)=\frac{2 x+5 y}{x}

\newline

Is

g(x)=x^{5}-2 x

a solution to the above equation?

\newline

Choose

1

answer:

\newline

(A) Yes

\newline

(B)

\mathrm{No}

Find Derivative of g(x): To determine if $g(x)=x^{5}-2x$ is a solution to the given derivative $g^{\prime}(x)=\frac{2x+5y}{x}$ , we need to find the derivative of $g(x)$ and compare it to the given derivative. $\newline$ First, let's find the derivative of $g(x)=x^{5}-2x$ . $\newline$ Using the power rule for differentiation, the derivative of $x^{5}$ is $5x^{4}$ , and the derivative of $-2x$ is $-2$ . $\newline$ So, the derivative of $g(x)$ is $g^{\prime}(x)=5x^{4}-2$ .
Compare Derivatives: Now, let's compare the derivative we found, $g'(x)=5x^{4}-2$ , with the given derivative $g'(x)=\frac{2x+5y}{x}$ . We can see that these two expressions are not the same. The given derivative has terms involving both $x$ and $y$ , and it is in the form of a fraction, whereas the derivative we found is a polynomial in $x$ only and does not involve $y$ .
Conclusion: Since the derivatives do not match, we can conclude that $g(x)=x^{5}-2x$ is not a solution to the given derivative $g^{\prime}(x)=\frac{2x+5y}{x}$ .

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