Application of derivatives.pdf
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👉 APPLICATION OF DERIVATIVES 👈
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[ Application of Derivatives ]
The side of an equilateral triangle is increasing at the rate of 2 cm/s. The rate at which area increases when the side is 10 is
The side of an equilateral triangle is increasing at the rate of 2 cm/s. The rate at which area increases when the side is 10 is
Anonymous Quiz
19%
10 cm2/s
42%
10/3 cm2/s
15%
√3 cm2/s
23%
10√3 cm2/s
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CONTINUITY AND DIFFERENTIADILITY.pdf
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👉 CONTINUITY AND
DIFFERENTIABILITY 👈
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DIFFERENTIABILITY 👈
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Q.The function f(x) = [ln(1+ax)-ln(1-bx)]/x, not defined at x=0. The value should be assigned to f at x=0, so that it is continuous at x =0, is
Anonymous Quiz
29%
a+b
34%
a-b
18%
b-a
19%
ln a + ln b
Answer: Option (b) g’(0) = cos(log 2)
Explanation:
Given: f(x) = |log 2 – sinx|
g(x) = f(f(x))
g’(x) = f’(f(x))f’(x)
Hence, g’(0) = f’(f(0))f’(0)
When x tends to 0, log 2 > sinx
Thus, f(x) = log 2 -sinx
f’(x) = -cos x
f’(0) = – cos 0 = -1
f’(log 2) = – cos (log 2)
⇒g’(0) = (-cos(log 2)) (-1) = cos (log 2)
Hence, option (b) g’(0) = cos (log 2) is the correct answer.
Explanation:
Given: f(x) = |log 2 – sinx|
g(x) = f(f(x))
g’(x) = f’(f(x))f’(x)
Hence, g’(0) = f’(f(0))f’(0)
When x tends to 0, log 2 > sinx
Thus, f(x) = log 2 -sinx
f’(x) = -cos x
f’(0) = – cos 0 = -1
f’(log 2) = – cos (log 2)
⇒g’(0) = (-cos(log 2)) (-1) = cos (log 2)
Hence, option (b) g’(0) = cos (log 2) is the correct answer.