#### To determine

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

#### Answer

The equation of the tangent line to the curve
y=tan(πx24) at
(1,1) is
y=πx−π+1.

#### Explanation

**Given:**

The function is
y=tan(πx24).

**Derivative Rule:**

**Chain Rule**

If *h* is differentiable at *x* and *g* is differentiable at
h(x), then the composite function
F=g∘h defined by
F(x)=g(h(x)) is differentiable at *x* and
F′ is given by the product

F′(x)=g′(h(x))⋅h′(x) (1)

**Formula used:**

The equation of the tangent line at
(x1,y1) is,
y−y1=m(x−x1). (2)

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

**Calculation:**

The derivative of
y is
dydx, which is obtained as follows,

dydx=ddx(y)=ddx(tan(πx24))

Apply the chain rule as shown in equation (1),

ddx(tan(πx24))=sec2(πx24)ddx(πx24)=sec2(πx24)(2πx4)=sec2(πx24)(πx2)

Therefore, the derivative of
y=tan(πx24) is
dydx=sec2(πx24)(πx2).

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

m=dydx|x=1=sec2(π(1)24)(π(1)2)=1cos2(π4)(π2)=1[cos(π4)]2(π2)

Simplify further as,

m=1[cos(π4)]2(π2)=1[12]2(π2) [∵cos(π4)=12]=2(π2)=π

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

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

(y−1)=π(x−1)y−1=πx−πy=πx−π+1

Therefore, the equation of the tangent line to the curve
y=tan(πx24) at
(1,1) is
y=πx−π+1.

#### To determine

**To sketch:** The graph of the curve and the tangent line.

#### Explanation

**Given:**

The equation of the curve is
y=tan(πx24).

The equation of the tangent line is
y=πx−π+1.

**Graph:**

Use the online graphing calculator to draw the graph of the functions as shown below in Figure 1.

From the Figure 1, it is observed that the equation of the tangent line touches on the curve
y=tan(πx24) at the point
(1,1)**.**