Asked by Brain box
find the first, second and third derivative of : sin3x^2?
Answers
Answered by
oobleck
just use the chain rule
y = sin(3x^2)
y' = cos(3x^2)(6x)
y" = -sin(3x^2)(6x)^2 + cos(3x^2)(6)
y"' = -cos(3x^2)(6x)^3 - sin(3x^2)(2*6x)(6) - sin(3x^2)(6x)(6)
I'll let you do the algebra. You can always check your results by using wolframalpha.com
Just type <u>3rd derivative sin(3x^2)</u>
y = sin(3x^2)
y' = cos(3x^2)(6x)
y" = -sin(3x^2)(6x)^2 + cos(3x^2)(6)
y"' = -cos(3x^2)(6x)^3 - sin(3x^2)(2*6x)(6) - sin(3x^2)(6x)(6)
I'll let you do the algebra. You can always check your results by using wolframalpha.com
Just type <u>3rd derivative sin(3x^2)</u>
Answered by
Oyinkonsola
Just look for the first,second and third derivative
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