55. Differentiate each trigonometric identity to obtain a new (or familiar) identity. (a) $\tan x=\frac{\sin x}{\cos x}$ (b) $\sec x=\frac{1}{\cos x}$ (c) $\sin x+\cos x=\frac{1+\cot x}{\csc x}$

Problem 55E

55. Differentiate each trigonometric identity to obtain a new (or familiar) identity.

(a) $\tan x=\frac{\sin x}{\cos x}$

(b) $\sec x=\frac{1}{\cos x}$

Calculus, James Stewart 8th edition
2. Derivatives
4. Derivatives of Trigonometric Functions
Problem no. 55E from 152

Step-by-Step Solution

To determine

To find: The derivative of tanx=sinxcosx.

Answer

The derivative of tanx=sinxcosx is sec2x=1cos2x_.

Explanation

Given:

The function is tanx=sinxcosx.

Derivative Rule:

Quotient rule: If f(x) and g(x) are differentiable function, then

ddx[f(x)g(x)]=g(x)ddx[f(x)]−f(x)ddx[g(x)][g(x)]2 (1)

Sum Rule: The sum rule for two functions f1(x) and f2(x) is given as follows.

ddx[g1(x)+g2(x)]=ddx[g1(x)]+ddx[g2(x)] (2)

Calculation:

Obtain the derivative of tanx=sinxcosx.

ddx(tanx)=ddx(sinxcosx)

Apply Quotient Rule as shown in equation (1),

ddx[tanx]=[cosx]ddx[sinx]−[sinx]ddx[cosx] [cosx]2 sec2x=[cosx][cosx]−[sinx][−sinx] [cosx]2=cos2x+sin2x[cosx]2sec2x=1 cos2x (Qcos2x+sin2x=1 )

Therefore, the derivative of tanx=sinxcosx is sec2x=1 cos2x_

To determine

To find: The derivative of secx=1cosx.

Answer

The derivative of secx=1cosx is secxtanx=sinxcos2x_.

Explanation

Given:

The function is secx=1cosx.

Calculation:

Obtain the derivative of secx=1cosx.

ddx(secx)=ddx(1cosx)

Apply Quotient Rule as shown in equation (1),

ddx[secx]=[cosx]ddx[1]−[1]ddx[cosx][cosx]2 secxtanx=[cosx][0]−[1][−sinx][cosx]2secxtanx=sinx[cosx]2secxtanx=sinx cos2x

Therefore, the derivative of secx=1cosx is secxtanx=sinxcos2x_.

To determine

To find: The derivative of sinx+cosx=1+cotxcscx.

Answer

The derivative of sinx+cosx=1+cotxcscx is cosx−sinx=−1+cotxcscx_

Explanation

Given:

The function is sinx+cosx=1+cotxcscx.

Calculation:

The differentiation of given function is,

ddx(sinx+cosx)=ddx(1+cotxcscx)

Apply Quotient Rule as shown in equation (1),

ddx[sinx+cosx]=[cscx]ddx[1+cosx]−[1+cosx]ddx[cscx][cscx]2

Apply Sum Rule as shown in equation (2),

ddx[sinx]+ddx[cosx]=[cscx]ddx[1+cotx]−[1+cotx]ddx[cscx][cscx]2[cosx]+[−sinx]=[cscx][0−csc2x]−[1+cotx][−cscxcotx][cscx]2cosx−sinx=−csc3x+[cscxcotx+cscxcot2x][cscx]2=cscx(−csc2x+[cotx+cot2x])[cscx]2

Simplify further, the derivative function becomes,

cosx−sinx=(−csc2x+cot2x+cotx)cscx=(−csc2x+cot2x)+cotxcscx=−1+cotxcscx (Q1+cot2x=csc2x) cosx−sinx=−1+cotxcscx

Therefore, the derivative of sinx+cosx=1+cotxcscx is cosx−sinx=−1+cotxcscx_.