## Calculus: Early Transcendentals 8th Edition

The equation of tangent line to the curve at $(1,0)$ is $$(l):y=-\frac{3}{4}x+\frac{3}{4}$$
$$y=\sqrt[4]x-x$$$$y=x^{1/4}-x$$ 1) Find the derivative of $y$ $$y'=\frac{d}{dx}(x^{1/4})-\frac{d}{dx}(x)$$ $$y'=\frac{1}{4}x^{-3/4}-1$$ $$y'=\frac{1}{4x^{3/4}}-1$$ $$y'=\frac{1}{4\sqrt[4]{x^3}}-1$$ 2) Find the slope of the tangent line to the curve at $(1,0)$, or in fact, $y'(1)$ $$y'(1)=\frac{1}{4\sqrt[4]{1^3}}-1$$$$y'(1)=\frac{1}{4}-1=-\frac{3}{4}$$ 3) The tangent line to the curve at $(1,0)$ would have the following equation: $$(l):y=y'(1)x+b$$$$y=-\frac{3}{4}x+b$$ Since $(1,0)$ lies in the tangent line $(l)$, we can use that point to find $b$. In detail, $$-\frac{3}{4}\times1+b=0$$$$-\frac{3}{4}+b=0$$$$b=\frac{3}{4}$$ In conclusion, the equation of tangent line to the curve at $(1,0)$ is $$(l):y=-\frac{3}{4}x+\frac{3}{4}$$