#### To determine

**To find: **Equation of tangent line and normal line to the given curve at given point.

#### Answer

Equation of tangent line is: y=2x+1

Equation of normal line is: y=-12x+1

#### Explanation

Differentiate the given curve and then plug the given point in it to get slope of tangent line. Then use point slope form of line to get equation of tangent line. Then calculate the negative reciprocal of slope of tangent line. It is slope of normal line. Again with this new slope find the equation of normal line using the slope point form.

**1) Formula:**

i. Power rule combined with chain rule:aa

ddxf(x)n=n.fxn-1.f'(x)

ii. Sum rule:

ddxfx+gx=ddxfx+ddxg(x)

iii. Constant multiple rule:

ddxk.fx=k.ddxfx where k is constant.

iv.

ddxsinx=cosx

v. Constant rule:

ddxk=0 where k is constant.

vi. Normal line and tangent lines are perpendicular to each other.

vii. Slopes of perpendicular lines are negative reciprocal of each other.

viii. Equation of line passing through point (a,f(a)) with slope m is, (y-f(a))=m(x-a)

**2) Given:**

y=1+4sinx, (0 ,1)

3) **Calculation:**

Differentiate y with respect to x,

f'(x)=ddx1+4sinx

By using power rule combined with chain rule,

ddx1+4sinx=121+4sinx.ddx(1+4sinx)

By using sum and constant multiple rule,

=121+4sinxddx1+4ddxsinx)

=4cosx21+4sinx

f'x=2cosx1+4sinx

At (0 ,1)

f'0=2cos01+4sin0=2

Use the point (0, 1) and this slope m = 2 in point slope form to get,

y-1=2*(x-0)

y-1=2x

adding 1 on both side,

y=2x+1

This is the required equation of tangent line at (0,1).

To get equation of normal line, find the negative reciprocal of slope of tangent line.

So, slope of normal line = -12

Using point slope form to get normal line,

y-1=-12*(x-0)

y= -12x+1

This is the required equation of normal line.

**Conclusion:**

Equation of tangent line is: y=2x+1

Equation of normal line is: y=-12x+1