Let's solve the given problem **step-by-step**. ### Given: \[ y = a^{x} \] Prove that: \[ \frac{dy}{dx} = \frac{y^2 \log y}{x(1-y)\log a} \] --- ### **Step 1: Start with the given equation** \[ y = a^x \] --- ### **Step 2: Take the derivative of both sides with respect to \(x\)** \[ \frac{dy}{dx} = a^x \cdot \log a \] But, \(a^x = y\), so: \[ \frac{dy}{dx} = y \log a \] --- ### **Step 3: Express \(a\) in terms of \(y\) and \(x\)** From \(y = a^x\): Take \(\log\) both sides (natural log): \[ \ln y = x \ln a \] \[ \ln a = \frac{\ln y}{x} \] --- ### **Step 4: Substitute \(\log a\) into the derivative** \[ \frac{dy}{dx} = y \cdot \frac{\ln y}{x} \] --- ### **Step 5: Express in terms of \(\log y\) (if base 10 is intended, use \(\log\); if not, stick with \(\ln\))** Assume "log" means natural logarithm, so: \[ \frac{dy}{dx} = \frac{y \ln y}{x} \] --- ### **Step 6: Manipulate to the required form** The required form is: \[ \frac{dy}{dx} = \frac{y^2 \log y}{x(1-y)\log a} \] Let's try to express \(y^2\) and \(1-y\): Recall: - \(y = a^x\) - So, \(1-y = 1-a^x\) - \(y^2 = (a^x)^2 = a^{2x}\) Let's manipulate: From earlier, \[ \frac{dy}{dx} = y \log a \] Multiply both numerator and denominator by \(y\): \[ \frac{dy}{dx} = y \log a = \frac{y^2 \log a}{y} \] But that's not matching the required form. Let's observe: Given: \[ \frac{dy}{dx} = \frac{y^2 \log y}{x(1-y)\log a} \] Let’s try to express \(\log y\) in terms of \(x\) and \(\log a\): Since \(y = a^x\), then \(\log y = x \log a\). So, \[ \frac{dy}{dx} = \frac{y^2 \cdot x \log a}{x (1-y) \log a} \] \[ = \frac{y^2 x \log a}{x (1-y) \log a} \] \[ = \frac{y^2}{1-y} \] But from earlier, \(\frac{dy}{dx} = y \log a\). So, unless there is a typo in the question, the correct derivative is: \[ \frac{dy}{dx} = y \log a \] But if you manipulate the given expression (by substituting \(\log y = x \log a\)), you get: \[ \frac{y^2 \log y}{x(1-y)\log a} = \frac{y^2 \cdot x \log a}{x(1-y)\log a} = \frac{y^2}{1-y} \] Which is **not** equal to \(y \log a\). --- ### **Final Answer** **The derivative of \(y = a^x\) is:** \[ \frac{dy}{dx} = y \log a = a^x \log a \] **If you substitute \(\log y = x \log a\) into the given expression, it simplifies to \(\frac{y^2}{1-y}\), which is NOT equal to \(y \log a\).** #### **Therefore, the given expression does not match the actual derivative unless there is some additional context or a typo in the problem statement.** --- #### **If you have a different function for \(y\), please provide it for further clarification.**