Скачать или смотреть Prove that d/dx (secx) = sec(x)tan(x)

Prove that d/dx (secx) = sec(x)tan(x)
We are asked to prove that:

d/dx (sec(x)) = sec(x)tan(x)

Step 1: Recall the definition of sec(x)
Secant is the reciprocal of cosine:

sec(x) = 1 / cos(x)

Step 2: Use the quotient rule
To differentiate sec(x) = 1 / cos(x), we apply the quotient rule for derivatives. The quotient rule states:

d/dx [f(x) / g(x)] = [g(x) * f'(x) - f(x) * g'(x)] / [g(x)]^2

Let:
f(x) = 1
g(x) = cos(x)

Now, differentiate f(x) and g(x):
f'(x) = 0
g'(x) = -sin(x)

Step 3: Apply the quotient rule
Using the quotient rule:

d/dx [sec(x)] = [cos(x) * 0 - 1 * (-sin(x))] / [cos(x)]^2

Simplify the expression:

d/dx [sec(x)] = sin(x) / cos(x)^2

Step 4: Simplify using trigonometric identities
We can now simplify sin(x) / cos(x)^2:

sin(x) / cos(x)^2 = (sin(x) / cos(x)) * (1 / cos(x))

Recall that:
sin(x) / cos(x) = tan(x)
1 / cos(x) = sec(x)

Thus:

d/dx [sec(x)] = sec(x) * tan(x)

This completes the proof.

Therefore:

d/dx (sec(x)) = sec(x)tan(x)