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Find an equation of the tangent line to the curve at the given point.

$ y = \sqrt[4]{x} - x, (1,0) $

$$y= \sqrt[4]{x}-x$$

First, rewrite $\sqrt[4]{x}$ in indices form using the property $\sqrt[a]{b}= b^{\frac{1}{a}}$

$$y= x^{\frac{1}{4}} -x$$

Recall, that the first derivative of an equation gives the equation of the gradient at any point $x$, therefore differenciate the equation with respect to $x$.

$$\frac{dy}{dx}=\frac{d}{dx}(x^{\frac{1}{4}})- \frac{d}{dx}(x)$$

Using the power rule:

$$\frac{dy}{dx}= \frac{1}{4}x^{-\frac{3}{4}}-1$$

To find the gradient, subctitute $x=1$

$$\frac{dy}{dx}= \frac{1}{4}(1)^{-\frac{3}{4}}-1$$

Rewrite using laws of indices:

$$\frac{dy}{dx}=\frac{1}{4}\sqrt[4]{(1)^{-3}}-1$$

Evaluate the gradient:

$$m=\frac{1}{4}(1)-1= \frac{-3}{4}$$

The point slope for is

$$y-y_1=m(x-x_1)$$

Substitute $m= -\frac{3}{4}$ and $y_1=0$ and $x_1=1$

$$y-0= -\frac{3}{4}(x-1)$$

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Baylor University

University of Michigan - Ann Arbor

University of Nottingham

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