$ latex u$ is the numerator as well as $latex v$ is the . As a result, we have

$ latex u = wrong {(x)} $
$ latex v =tan {(x)} $
$ latex f( x) = frac {u} {v} $

Next, we acquire $latex u$ as well as $latex v$ separately:

$ latex u = wrong {(x)} $
$ latex u’ = cos {(x)} $

$ latex v= tan {(x)} $
$ latex v’ = sec ^ {2} {(x)} $

Now, we can make use of the ratio policy formula to acquire our provided trouble:

$$ frac {d} {dx} (frac {u} {v}) = frac {vu’- uv’} {v ^ 2} $$

$$ frac {d} {dx} f( x) = frac {(wrong {(x)}) cdot (sec ^ {2} {(x)})– (tan {(x)}) cdot (cos {(x)})} {(tan {(x)} )^ 2} $$

Simplifying algebraically as well as using trigonometric identifications, we get

$$ frac {d} {dx} f( x) = frac {(wrong {(x)}) cdot (frac {1} {cos ^ {2} {(x)}})– (frac {wrong {(x)}} {cos {(x)}}) cdot (cos {(x)})} {tan ^ {2} {(x)}} $$

$$ frac {d} {dx} f( x) = frac {(frac {wrong {(x)}} {cos {(x)}}) cdot (frac {1} {cos {(x)}})– (wrong {(x)}} {tan ^ {2} {(x)}} $$

$$ frac {d} {dx} f( x) = frac {tan {(x)} cdot sec {(x)}– wrong {(x)}} {tan ^ {2} {(x)}} $$

$$ frac {d} {dx} f( x) = frac {tan {(x)} sec {(x)}} {tan ^ {2} {(x)}}– frac {wrong {(x)}} {tan ^ {2} {(x)}} $$

$$ frac {d} {dx} f( x) = frac {tan {(x)} sec {(x)}} {tan ^ {2} {(x)}}– frac {wrong {(x)}} {(frac {wrong {(x)}} {cos {(x)}} )^ 2} $$

$$ frac {d} {dx} f( x) = frac {sec {(x)}} {tan {(x)}}– frac {wrong {(x)} cos ^ {2} {(x)}} {wrong ^ {2} {(x)}} $$

$$ frac {d} {dx} f( x) = left[frac{frac{1}{cos{(x)}}}{frac{cos{(x)}}{sin{(x)}}}right]– left[sin{(x)} cdot frac{cos^{2}{(x)}}{sin^{2}{(x)}}right]$$

$$ frac {d} {dx} f( x) = frac {cos {(x)}} {cos {(x)} cdot wrong {(x)}}– frac {cos ^ {2} {(x)}} {wrong {(x)}} $$

$$ frac {d} {dx} f( x) = frac {1} {wrong {(x)}}– frac {wrong ^ {2} {(x)} -1} {wrong {(x)}} $$

$$ frac {d} {dx} f( x) = frac {1– wrong ^ {2} {(x)}– 1} {wrong {(x)}} $$

$$ frac {d} {dx} f( x) = frac {-wrong ^ {2} {(x)}} {wrong {(x)}} $$

And the last response is:

$ latex f'( x) = -wrong {(x)} $



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