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Chapter 2: Q 19. (page 238)

Find the derivatives of the function: $f (x) = \ln (\tan^{2} (x)) .$

Short Answer

Expert verified

The derivative of $f (x) = \ln (\tan^{2} (x)) i s \frac{2}{\sin (x) \cos (x)} .$

Step by step solution

Given information

The given expression is $f (x) = \ln (\tan^{2} (x)) .$

Simplification

$f' (x) = \frac{d}{d x} (\ln (\tan^{2} (x)) = \frac{1}{\tan^{2} (x)} \times 2 \tan (x) s e c^{2} (x) = \frac{2}{\sin (x) \cos (x)} H e n c e, f' (x) = \frac{2}{\sin (x) \cos (x)} .$

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Most popular questions from this chapter

Use the definition of the derivative to find the equations of the lines described in Exercises 59-64.The line tangent to the graph of $y = 1 - x - x^{2}$ at the point $(1, - 1)$

Think about what you did today and how far north you were from your house or dorm throughout the day. Sketch a graph that represents your distance north from your house or dorm over the course of the day, and explain how the graph reflects what you did today. Then sketch a graph of your velocity.

Use (a) the $h \to 0$ definition of the derivative and then

(b) the $z \to c$ definition of the derivative to find $f' (c)$ for each function f and value $x = c$ in Exercises 23–38.

27. $f (x) = 1 - x^{3}, x = - 1$

Use (a) the $h \to 0$ definition of the derivative and then

(b) the $z \to c$ definition of the derivative to find $f' (c)$ for each function f and value $x = c$ in Exercises 23–38.

23. $f (x) = x^{2}, x = - 3$

Find a function that has the given derivative and value. In each case you can find the answer with an educated guess and check process it may be helpful to do some preliminary algebra

$f' (x) = (x^{4} - 8) (1 - 3 x^{5}); f (0) = 2$

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Find the derivatives of the function: f(x)=lntan2x.

Short Answer

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Find the derivatives of the function: $f (x) = \ln (\tan^{2} (x)) .$