differentiate tan 3x+4
Answers
Step-by-step explanation:
lim
θ
tanθ
=1
\displaystyle\tan(A+B) = \frac{\tan A+\tan B}{1-\tan A\tan B}tan(A+B)=
1−tanAtanB
tanA+tanB
Then
\begin{gathered}\displaystyle\frac{d}{dx}\tan(3x+4) = \lim_{h\rightarrow0}\frac1h\left[\tan(3(x+h)+4)-\tan(3x+4)\right] \\ \\=\lim_{h\rightarrow0}\frac1h\left[\frac{\tan(3x+4)+\tan(3h)}{1-\tan(3x+4)\tan(3h)}-\tan(3x+4)\right] \\ \\=\lim_{h\rightarrow0}\frac1h\left[\frac{\tan(3x+4)+\tan(3h)-\tan(3x+4)+\tan^2(3x+4)\tan(3h)}{1-\tan(3x+4)\tan(3h)}\right] \\ \\=\lim_{h\rightarrow0}\frac{\tan(3h)}h\left[\frac{1+\tan^2(3x+4)}{1-\tan(3x+4)\tan(3h)}\right]\end{gathered}
dx
d
tan(3x+4)=
h→0
lim
h
1
[tan(3(x+h)+4)−tan(3x+4)]
=
h→0
lim
h
1
[
1−tan(3x+4)tan(3h)
tan(3x+4)+tan(3h)
−tan(3x+4)]
=
h→0
lim
h
1
[
1−tan(3x+4)tan(3h)
tan(3x+4)+tan(3h)−tan(3x+4)+tan
2
(3x+4)tan(3h)
]
=
h→0
lim
h
tan(3h)
[
1−tan(3x+4)tan(3h)
1+tan
2
(3x+4)
]
\begin{gathered}\displaystyle=\lim_{h\rightarrow0}\frac{\tan(3h)}h\left[\frac{1+\tan^2(3x+4)}{1-\tan(3x+4)\tan(3h)}\right] \\ \\=\lim_{h\rightarrow0}3\times\frac{\tan(3h)}{3h}\times\frac{\sec^2(3x+4)}{1-\tan(3x+4)\tan(3h)}\\ \\= 3\times 1\times \frac{\sec^2(3x+4)}{1-0}\\ \\=3\sec^2(3x+4)\end{gathered}
=
h→0
lim
h
tan(3h)
[
1−tan(3x+4)tan(3h)
1+tan
2
(3x+4)
]
=
h→0
lim
3×
3h
tan(3h)
×
1−tan(3x+4)tan(3h)
sec
2
(3x+4)
=3×1×
1−0
sec
2
(3x+4)
=3sec
2
(3x+4)
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