#### To determine

**(a)**

**To find:**f'and f".

#### Answer

f'x=4-sec2x

f"(x)=-2sec2xtanx

#### Explanation

**1) Formula:**

i. difference rule:

ddxfx-gx=ddxfx-ddxg(x)

ii. Power rule combined with chain rule:

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

iii. constant function rule:

ddxc=0 where c is constant.

iv.Equation of tangent line to the curve y=f(x) at point (a,f(a)) is,

(y-f(a))=f’(a)(x-a)

**3) Given:**

y=fx=4x-tanx , -π2<c<π2

**4) Calculation:**

Differentiate given equation with respect to x,

f'x=ddx[4x-tanx]

Using product rule,

ddx4x-tanx=ddx4x-ddxtanx=4-sec2x

f'x=4-sec2x

Again differentiate f’ with respect to x,

f"(x)=ddxf'x=ddx4-sec2x=0-2.secxddxsecx=-2secx.secxtanx=-2sec2xtanx

**Final statement:**

f'x=4-sec2x

f"(x)=-2sec2xtanx

#### To determine

**(b)**

**To check:** answer to part(a) is reasonable by comparing graphs of f,f’,f”

#### Answer

Answers from part a are reasonable by comparing graphs of f(x), f’x and f''(x).

#### Explanation

By using a graphing utility the graphs of (x), f’x and f''(x) are

Here blue graph is for f(x), black graph is for f'(x) and red graph is for f''(x).

From these two graphs, notice that f'(x) is positive when graph of f(x) is increasing and negative when graph of f(x) is decreasing.

f''(x) is positive when graph of f'(x) is increasing and negative when graph of f'(x) is decreasing.

**Final statement:**

Answers from part a are reasonable by comparing graphs of (x), f’x and f''(x).