Given that f(x) = x ( x), find f'(x) when x = pi/4.

Question

ScanSolve AI Tutor · Accepted Answer

It seems there was some confusion with the numbering in our previous exchanges, as the image you sent contains questions 1 through 5. I have already solved Question 1 (parts a and b) and Question 4 ii) from this image, which I previously referred to as "Question 8" and "Question 7" respectively.

Let's continue by solving Question 3 from the image.

Question 3: Given that $f(x) = \sin x \ln(\tan x)$ , find $f'(x)$ when $x = \frac{\pi}{4}$ .

Identify the function and the required operation.
We need to find the derivative $f'(x)$ of $f(x) = \sin x \ln(\tan x)$ using the product rule, and then evaluate it at $x = \frac{\pi}{4}$ .
The product rule states that if $f(x) = u(x)v(x)$ , then $f'(x) = u'(x)v(x) + u(x)v'(x)$ .
Define $u(x)$ and $v(x)$ and find their derivatives.
Let $u(x) = \sin x$ .
Then $u'(x) = \cos x$ .

Let $v(x) = \ln(\tan x)$ .
To find $v'(x)$ , we use the chain rule. Let $w = \tan x$ , so $v(x) = \ln w$ .
$\frac{dv}{dw} = \frac{1}{w} = \frac{1}{\tan x}$ .
$\frac{dw}{dx} = \sec^2 x$ .
So, $v'(x) = \frac{dv}{dw} \cdot \frac{dw}{dx} = \frac{1}{\tan x} \cdot \sec^2 x$ .
We can simplify this:
$v'(x) = \frac{\cos x}{\sin x} \cdot \frac{1}{\cos^2 x} = \frac{1}{\sin x \cos x}$
Apply the product rule to find $f'(x)$ .
$f'(x) = u'(x)v(x) + u(x)v'(x)$ $f'(x) = (\cos x)(\ln(\tan x)) + (\sin x)\left(\frac{1}{\sin x \cos x}\right)$ $f'(x) = \cos x \ln(\tan x) + \frac{1}{\cos x}$ $f'(x) = \cos x \ln(\tan x) + \sec x$
Evaluate $f'(x)$ at $x = \frac{\pi}{4}$ .
Substitute $x = \frac{\pi}{4}$ into the expression for $f'(x)$ :
$f'\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) \ln\left(\tan\left(\frac{\pi}{4}\right)\right) + \sec\left(\frac{\pi}{4}\right)$
Recall the trigonometric values:
$\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$
$\tan\left(\frac{\pi}{4}\right) = 1$
$\sec\left(\frac{\pi}{4}\right) = \frac{1}{\cos\left(\frac{\pi}{4}\right)} = \frac{1}{\sqrt{2}/2} = \frac{2}{\sqrt{2}} = \sqrt{2}$
Also, $\ln(1) = 0$ .

Substitute these values:
$f'\left(\frac{\pi}{4}\right) = \left(\frac{\sqrt{2}}{2}\right) \cdot (0) + \sqrt{2}$ $f'\left(\frac{\pi}{4}\right) = 0 + \sqrt{2}$ $f'\left(\frac{\pi}{4}\right) = \sqrt{2}$
The value of $f'(x)$ when $x = \frac{\pi}{4}$ is $\boxed{\sqrt{2}}$ .

Given that f(x) = x ( x), find f'(x) when x = (pi)/(4).

Answer

Step-by-step solution

Identify the function and the required operation.

Define $u(x)$ and $v(x)$ and find their derivatives.

Apply the product rule to find $f'(x)$ .

Evaluate $f'(x)$ at $x = \frac{\pi}{4}$ .

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