Question

In Exercises \(33-36,\) find \(d y / d x\) $$y=x^{1+\sqrt{2}}(1+\sqrt{2}) x^{\sqrt{2}}$$

Short answer

The derivative of the function \(y = x^{1+\sqrt{2}}(1+\sqrt{2}) x^{\sqrt{2}}\) is \(dy/dx = 4x^{3\sqrt{2}}-x\)

Step-by-step solution

Identify the two functions

The given function can be written as a product of two functions: \(u(x) = x^{1+\sqrt{2}}\) and \(v(x) = (1+\sqrt{2}) x^{\sqrt{2}}\). We will differentiate them separately.

Differentiate the first function

Using the power rule for differentiation, which states that the derivative of \(x^n\) is \(n*x^{n-1}\), the derivative of the first function \(u(x)\) is \(du/dx = (1 + \sqrt{2}) * x^{\sqrt{2}}\).

Differentiate the second function

The second function is itself a product of two functions: \(a(x) = (1+\sqrt{2})\) and \(b(x) = x^{\sqrt{2}}\). So, we have to use the product rule again. \[ \rac{dv}{dx} = a'(x)b(x) + a(x)b'(x) \\ \rac{dv}{dx} = 0*x^{\sqrt{2}} + (1+\sqrt{2})*x^{\sqrt{2}}*(\sqrt{2}*x^{\sqrt{2}-1}) \\ \rac{dv}{dx} = 2(1+\sqrt{2})*x^{2*\sqrt{2}-1} \]

Apply the product rule

Finally, we can put these pieces together using the product rule. The derivative of the original function is \[ \rac{dy}{dx} = u'(x)v(x) + u(x)v'(x) \\ \rac{dy}{dx} = (1 + \sqrt{2}) x^{\sqrt{2}} * (1+\sqrt{2}) x^{\sqrt{2}} + x^{1+\sqrt{2}} * 2(1+\sqrt{2})*x^{2\sqrt{2}-1} \\ \rac{dy}{dx} = 2x^{3\sqrt{2}} + 2x^{3\sqrt{2}}-x \\ \rac{dy}{dx} = 4x^{3\sqrt{2}}-x \]