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Problem 36 Easy Difficulty

Find an equation of the tangent line to the curve at the given point.
$ y = \sqrt[4]{x} - x, (1,0) $

Answer

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

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