Determining the Slope of a Tangent to a Curve Involving Radicals
<p>Given \( f(x) = x^{\frac{7}{5}} + \frac{4}{\sqrt[5]{x^3}} \), to find the slope of the tangent line at any point \( x \), we need to find the derivative of \( f(x) \) with respect to \( x \).</p>
<p>First, we rewrite the function in a more derivative-friendly form:</p>
<p>\( f(x) = x^{\frac{7}{5}} + 4x^{-\frac{3}{5}} \)</p>
<p>Now we differentiate \( f(x) \) with respect to \( x \) using the power rule \( \frac{d}{dx} x^n = nx^{n-1} \):</p>
<p>\( f'(x) = \frac{7}{5}x^{\frac{7}{5}-1} - \frac{12}{5}x^{-\frac{3}{5}-1} \)</p>
<p>Simplifying, we get:</p>
<p>\( f'(x) = \frac{7}{5}x^{\frac{2}{5}} - \frac{12}{5}x^{-\frac{8}{5}} \)</p>
<p>So, the slope of the tangent line to the curve at any point \( x \) is given by \( f'(x) \).</p>
