प्रथम सिद्धांत से निम्नलिखित फलनों के अवकलज ज्ञात कीजिए :
Answers
Step-by-step explanation:
Hence, the general solution of sin θ = 1 is θ = (4n + 1)π2, n ∈ Z. Therefore, either, 2 sin x + 3 = 0 ⇒ sin x = - 32, Which is impossible since the numerical value of sin x cannot be greater than 1. We know that the general solution of sin θ = 1 is θ = (4n + 1)π2, n ∈ Z.
f'(x) = Cos(x + 1) यदि f(x) = Sin(x + 1)
Step-by-step explanation:
प्रथम सिद्धांत
f'(x) = Lim h → 0 (f(x + h) - f(x) )/h
f(x) = Sin(x + 1)
f'(x) = Lim h → 0 (Sin(x + h + 1) - Sin(x + 1) )/h
=> f'(x) = Lim h → 0 (Sin(x + 1 + h) - Sin(x + 1) )/h
Sin(A + B) = SinACosB + CosASinB
=> f'(x) = Lim h → 0 (Sin(x + 1)Cosh + Cos(x + 1)Sinh - Sin(x + 1) )/h
=> f'(x) = Lim h → 0 (Sin(x + 1)Cosh - Sin(x + 1) )/h + Cos(x + 1)Sinh/h
Lim h → 0 Sinh/h = 1
=> f'(x) = Lim h → 0 (Sin(x + 1)Cosh - Sin(x + 1) )/h + Cos(x + 1)
=> f'(x) = Sin(x + 1)Cos(0) - Sin(x +1) + Cos(x + 1)
=> f'(x) = Sin(x + 1) - Sin(x +1) + Cos(x + 1)
=> f'(x) = Cos(x + 1)
f'(x) = Cos(x + 1) यदि f(x) = Sin(x + 1)
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