Chapter 2: Q. 64 (page 234)

Use logarithmic differentiation to find derivative
$f (x) = (\sec x)^{x}$

Short Answer

Expert verified

The answer of this question is

$f^{'} (x) = (\ln (\sec x) + x \tan x) (\sec x)^{x}$

Step by step solution

Step 1.Given information

We have been given to find the derivative of function

$f (x) = (\sec x)^{x}$

Step 2.Derivative of given function

Product rule

$(f g)^{'} (x) = f^{'} (x) g (x) + f (x) g^{'} (x)$

Quotient rule

${(\frac{f}{g})}^{'} (x) = \frac{f^{'} (x) g (x) - f (x) g^{'} (x)}{(g (x))^{2}}$

Step 3.Using logarithmic approach

Taking log both side and then approaching

$\begin{aligned} f (x) = (\sec x)^{x} \\ \ln f (x) = \ln (\sec x)^{x} [Taking log in both sides] \\ \ln f (x) = x \ln (\sec x) \\ \frac{f^{'} (x)}{f (x)} = \frac{d}{d x} (x \ln (\sec x)) [\begin{matrix} Differentiate both sides \\ with respect to x \end{matrix}] \\ \frac{f^{'} (x)}{f (x)} = (\frac{d}{d x} (x)) \ln (\sec x) + x (\frac{d}{d x} \ln (\sec x)) \\ = \ln (\sec x) + \frac{x}{\sec x} (\frac{d}{d x} \sec x) [\begin{matrix} Using product rule \\ \frac{d}{d x} \ln x = \frac{1}{x} \end{matrix}] \\ = \ln (\sec x) + \frac{x}{\sec x} (\sec x \tan x) \\ f^{'} (x) = (\frac{d}{d x} \ln (\sec x = \sec x \tan x] \end{aligned}]$

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Use logarithmic differentiation to find derivative
$f (x) = (\sec x)^{x}$

Short Answer

Step by step solution

Step 1.Given information

Step 2.Derivative of given function

Step 3.Using logarithmic approach

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Use logarithmic differentiation to find derivativef(x)=(sec⁡x)x

Short Answer

Step by step solution

Step 1.Given information

Step 2.Derivative of given function

Step 3.Using logarithmic approach

One App. One Place for Learning.

Most popular questions from this chapter

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Use logarithmic differentiation to find derivative
$f (x) = (\sec x)^{x}$