Question

question 7 consider two functions, f(x) and g(x), related as follows: g(ln(x))f(x)=a cos(x), where a is a constant. which of the following statements is true? -a sin(x)=g(ln(x))f(x)+g(ln(x))f(x) -a sin(x)=g($\frac{1}{x}$)f(x)+g(ln(x))f(x) f(x)=$\frac{\frac{1}{\sqrt{1 - x}\sqrt{1 + x}}+\frac{g(x)f(x)}{g(x)}}{g(ln(x))}$ f(x)=-\frac{g(ln(x))f(x)+ax sin(x)}{xg(ln(x))} 0 = \frac{g(x)f(x)}{x}+g(ln(x))f(x)+a sin(x)

Explanation:

Step1: Differentiate both sides

Differentiate $g(\ln(x))f(x)=A\cos(x)$ using the product - rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u = g(\ln(x))$ and $v = f(x)$. The derivative of the left - hand side is $g(\ln(x))f^\prime(x)+g^\prime(\ln(x))\frac{1}{x}f(x)$ by the chain rule (since if $y = g(u)$ and $u=\ln(x)$, then $\frac{dy}{dx}=g^\prime(u)\frac{du}{dx}=g^\prime(\ln(x))\frac{1}{x}$), and the derivative of the right - hand side is $-A\sin(x)$. So, $g(\ln(x))f^\prime(x)+g^\prime(\ln(x))\frac{1}{x}f(x)=-A\sin(x)$.

Answer:

The first option $-A\sin(x)=g(\ln(x))f^\prime(x)+g^\prime(\ln(x))f(x)$ is incorrect because of the missing $\frac{1}{x}$ factor in the second term on the right - hand side. The second option $-A\sin(x)=g^\prime(\frac{1}{x})f(x)+g(\ln(x))f^\prime(x)$ is incorrect as the argument of $g^\prime$ should be $\ln(x)$ with the correct chain - rule factor. The third and fourth options are incorrect as they do not follow from the correct application of the product and chain rules. The correct statement is $-A\sin(x)=g(\ln(x))f^\prime(x)+g^\prime(\ln(x))\frac{1}{x}f(x)$. But among the given options, none of them are exactly correct based on the proper differentiation. If we assume there is a mis - typing in the options and we focus on the form of the product and chain rule application, the closest one is the first option with the caveat of the missing $\frac{1}{x}$ factor. However, if we consider the problem as it is presented, there is no correct option among the ones provided.