y=sin(ax+b)then dy÷dx?explain
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y = sin (ax + b)
dy/dx = d/dx sin (ax + b)
= cos (ax + b)* d/dx ( ax + b) [ d/dx sin ∅ = cos ∅]
= cos (ax + b) ( a + 0) [d/dx ax = a and d/dx b = 0]
= a cos (ax + b)
d/dx log cot x = 1/cot x * d/dx cot x [d/dx log x = 1/x]
= 1/cot x * - cosec² x [ d/dx cot ∅ = - cosec² ∅ ]
= - cosec² x/cot x
dy/dx = d/dx sin (ax + b)
= cos (ax + b)* d/dx ( ax + b) [ d/dx sin ∅ = cos ∅]
= cos (ax + b) ( a + 0) [d/dx ax = a and d/dx b = 0]
= a cos (ax + b)
d/dx log cot x = 1/cot x * d/dx cot x [d/dx log x = 1/x]
= 1/cot x * - cosec² x [ d/dx cot ∅ = - cosec² ∅ ]
= - cosec² x/cot x
Anonymous:
ur welcome
y=log cotx then dy÷dx ?explain
d/dx log(cot x)
1/cot x * d/dx (cot x)
1/cot x * -sin x = - sinx/cot x
the ans is - tan x [sin x/cos x = tan x]
how send clearly
I will add it in my answer
I added it in my answer
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i can not understand your question
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