sin (4. Sinx +5.cosx √41
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Answer:
Let y=sin
−1
(
41
4sinx+5cosx
)
=sin
−1
[(sinx)(
41
4
)+(cosx)(
41
5
)]
since, (
41
4
)
2
+(
41
5
)
2
=
41
16
+
41
25
=1
we can write,
41
4
=cos∞ and
41
5
=sin∞
∴y=sin
−1
(sinxcos∞+cosxsin∞)
=sin
−1
[sin(x+∞)]
=x+∞, where ∞ is a constant
Differentiating w.r.t. x, we get
dx
dy
=
dx
d
(x+∞)
=
dx
d
(x)+
dx
d
(∞)
=1+0
=1
Step-by-step explanation:
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