d/dx[log (tan4-3x)]= ?
Answers
Answer:
SOLUTION
GIVEN
y = log [ tan ( 4 - 3x )]
TO DETERMINE
\displaystyle \sf{ \frac{dy}{dx} }
dx
dy
EVALUATION
Here it is given that
y = log [ tan ( 4 - 3x )]
Now Differentiating both sides with respect to x we get
\displaystyle \sf{ \frac{dy}{dx} = \frac{d}{dx} \log \tan(4 - 3x)}
dx
dy
=
dx
d
logtan(4−3x)
\displaystyle \sf{ \implies \frac{dy}{dx} = \frac{1}{\tan(4 - 3x)} \frac{d}{dx} \tan(4 - 3x)}⟹
dx
dy
=
tan(4−3x)
1
dx
d
tan(4−3x)
\displaystyle \sf{ \implies \frac{dy}{dx} = \frac{1}{\tan(4 - 3x)} \times { \sec}^{2} (4 - 3x) \times \frac{d}{dx} (4 - 3x)}⟹
dx
dy
=
tan(4−3x)
1
×sec
2
(4−3x)×
dx
d
(4−3x)
\displaystyle \sf{ \implies \frac{dy}{dx} = \frac{1}{\tan(4 - 3x)} \times { \sec}^{2} (4 - 3x) \times - 3}⟹
dx
dy
=
tan(4−3x)
1
×sec
2
(4−3x)×−3
\displaystyle \sf{ \implies \frac{dy}{dx} = - 3 \times \frac{{ \sec}^{2} (4 - 3x)}{\tan(4 - 3x)} }⟹
dx
dy
=−3×
tan(4−3x)
sec
2
(4−3x)
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Step-by-step explanation: