Compute for the derivative of the given function for \( x=-4 \). \[ y=\frac{\sqrt{x^{2}+4}}{8 x} \] niwer: nswe uestion 7 :ompute for the derivative of the given function for \( \mathrm{x}=-4 \) \[ =\left(x^{2}-7\right) \sqrt{x^{2}-5} \] swer

We start with the two functions. In each case our goal is to differentiate the function and then evaluate the derivative at \( x=-4 \). --- ### 1. For the function \[ y=\frac{\sqrt{x^2+4}}{8x}, \] we proceed as follows. Write \[ y(x)=\frac{f(x)}{g(x)} \quad \text{with} \quad f(x)=\sqrt{x^2+4}=(x^2+4)^{\frac{1}{2}} \quad \text{and} \quad g(x)=8x. \] The quotient rule gives \[ y'(x)=\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}. \] 1. **Differentiate \( f(x) \):** \[ f'(x)=\frac{d}{dx}\,(x^2+4)^{\frac{1}{2}}=\frac{1}{2}(x^2+4)^{-\frac{1}{2}}\cdot 2x=\frac{x}{\sqrt{x^2+4}}. \] 2. **Differentiate \( g(x) \):** \[ g'(x)=\frac{d}{dx}(8x)=8. \] 3. **Apply the quotient rule:** \[ y'(x)=\frac{\frac{x}{\sqrt{x^2+4}}(8x)-\sqrt{x^2+4}(8)}{(8x)^2}. \] Simplify the numerator: \[ \frac{8x^2}{\sqrt{x^2+4}}-8\sqrt{x^2+4} = 8\left(\frac{x^2-\left(x^2+4\right)}{\sqrt{x^2+4}}\right)=\frac{-32}{\sqrt{x^2+4}}. \] The denominator is \[ (8x)^2=64x^2. \] Hence, \[ y'(x)=\frac{-32}{64x^2\sqrt{x^2+4}}=\frac{-1}{2x^2\sqrt{x^2+4}}. \] 4. **Evaluate at \( x=-4 \):** Compute \[ (-4)^2=16,\quad \sqrt{16+4}=\sqrt{20}=2\sqrt{5}. \] So, \[ y'(-4)=\frac{-1}{2\cdot 16\cdot2\sqrt{5}}=\frac{-1}{64\sqrt{5}}. \] Thus, the derivative at \( x=-4 \) is \[ y'(-4)=-\frac{1}{64\sqrt{5}}. \] --- ### 2. For the function \[ y=(x^2-7)\sqrt{x^2-5}, \] we use the product rule. Let \[ u(x)=x^2-7\quad \text{and} \quad v(x)=\sqrt{x^2-5}=(x^2-5)^{\frac{1}{2}}. \] Then \[ y(x)=u(x)v(x) \] and the product rule gives \[ y'(x)=u'(x)v(x)+u(x)v'(x). \] 1. **Differentiate \( u(x) \):** \[ u'(x)=\frac{d}{dx}(x^2-7)=2x. \] 2. **Differentiate \( v(x) \):** \[ v'(x)=\frac{d}{dx} (x^2-5)^{\frac{1}{2}}=\frac{1}{2}(x^2-5)^{-\frac{1}{2}}\cdot 2x=\frac{x}{\sqrt{x^2-5}}. \] 3. **Write the derivative:** \[ y'(x)=2x\sqrt{x^2-5}+(x^2-7)\frac{x}{\sqrt{x^2-5}}. \] 4. **Evaluate at \( x=-4 \):** Calculate each part: - For \( x=-4 \): \[ (-4)^2=16. \] - Then, \[ \sqrt{x^2-5}=\sqrt{16-5}=\sqrt{11}. \] - Also, \[ x^2-7=16-7=9. \] Hence, the first term becomes \[ 2x\sqrt{x^2-5}=2(-4)\sqrt{11}=-8\sqrt{11}, \] and the second term is \[ (x^2-7)\frac{x}{\sqrt{x^2-5}}=9\frac{-4}{\sqrt{11}}=-\frac{36}{\sqrt{11}}. \] Combining, \[ y'(-4)=-8\sqrt{11}-\frac{36}{\sqrt{11}}. \] We can also combine the two terms over a common denominator: \[ -8\sqrt{11} = -\frac{8\cdot11}{\sqrt{11}}=-\frac{88}{\sqrt{11}}, \] so that \[ y'(-4)=-\frac{88}{\sqrt{11}}-\frac{36}{\sqrt{11}}=-\frac{124}{\sqrt{11}}. \] Thus, the derivative at \( x=-4 \) is \[ y'(-4)=-\frac{124}{\sqrt{11}}. \] --- ### Final Answers 1. For \( y=\frac{\sqrt{x^2+4}}{8x} \): \[ y'(-4)=-\frac{1}{64\sqrt{5}}. \] 2. For \( y=(x^2-7)\sqrt{x^2-5} \): \[ y'(-4)=-8\sqrt{11}-\frac{36}{\sqrt{11}} \quad \text{or} \quad y'(-4)=-\frac{124}{\sqrt{11}}. \]