#### To determine

**To find:** The equation of the tangent line to the given equation at the point.

#### Answer

The equation of the tangent line to the equation y2=5x4−x2 at the point (1,2) is y=92x−52.

#### Explanation

**Given:**

The curve with equation y2=5x4−x2.

The point is (1,2).

**Derivative rules:**

(1) Chain rule: If y=f(u) and u=g(x) are both differentiable functions, then dydx=dydu⋅dudx.

(2) Product rule: If y=f(u) and u=g(x) are both differentiable function, then ddx(f(x)+g(x))=ddx(f(x))+ddx(g(x)).

**Formula used:**

The equation of the tangent line at (x1,y1) is, y−y1=m(x−x1) (1)

where, *m* is the slope of the tangent line at (x1,y1) and m=dydx|x=x1,y=y1.

**Calculation:**

Obtain the equation of tangent line to the given point.

y2=5x4−x2

Differentiate the above equation implicitly with respect to *x*,

ddx(y2)=ddx(5x4−x2)ddx(y2)=ddx(5x4)−ddx(x2)ddx(y2)=5ddx(x4)−ddx(x2)ddx(y2)=20x3−2x

Apply the chain rule (1) and simplify the terms,

ddy(y2)dydx=20x3−2x2ydydx=20x3−2xdydx=20x3−2x2ydydx=10x3−xy

Therefore, the derivative of the equation is dydx=10x3−xy.

The slope of the tangent line at (1,2) is computed as follows,

m=dydx|(x1,y1)=(1,2)=10(1)3−12=92

Thus, the slope of the tangent line at (1,2) is m=92.

Substitute (1,2) for (x1,y1) and m=92 in equation (1),

y−2=92(x−1)y−2=9x2−92y=9x2−92+2y=92x−52

Therefore, the equation of the tangent line to the equation y2=5x4−x2 at the point (1,2) is y=92x−52.

#### To determine

**To sketch:** The curve and tangent line to the given point.

#### Explanation

**Graph:**

Use online graphic calculator to draw the curve and the tangent line as shown below in Figure 1.

From Figure 1, it is observed that the line y=92x−52 touch the curve y2=5x4−x2 at (1,2). That is, the tangent line y=92x−52 to the curve at the point (1,2).