y = √x (x^3 + 1)
dy/dx = x(1/2) (3x^2) + (1/2)x^(-1/2) (x^3 + 1)
= (1/2)(x^(-1/2)) (6x^3 + x^3 + 1)
= (7x^3 + 1)/(2√x)
you are correct, but you need brackets around the denominator, or else it is only divided by 2
Find the derivative.
d/dx[sqrt of x (x^3 +1)]
My answer in its most simplified form:
(7x^3 + 1)/ 2 • sqrt of x
2 answers
If this is y = [ x^4 + x ]^.5 ??????
dy/dx = .5 [ x^4 + x ]^-.5 (4x^3 +1)
= .5 (4x^3+1) /sqrt(x^4+x)
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if you mean
y =( x^.5 )(x^3+1)
y = x^3.5 + x^.5
dy/dx = 3.5 x^2.5 + .5 x^.5 /x
= x^.5 ( 7 x^2 + 1/x)/2
= x^.5 (7 x^3 + 1) /(2 x)
= (7 x^3 + 1) /[ 2 sqrt x ]
agree, but parentheses are IMPORTANT
dy/dx = .5 [ x^4 + x ]^-.5 (4x^3 +1)
= .5 (4x^3+1) /sqrt(x^4+x)
=================================
if you mean
y =( x^.5 )(x^3+1)
y = x^3.5 + x^.5
dy/dx = 3.5 x^2.5 + .5 x^.5 /x
= x^.5 ( 7 x^2 + 1/x)/2
= x^.5 (7 x^3 + 1) /(2 x)
= (7 x^3 + 1) /[ 2 sqrt x ]
agree, but parentheses are IMPORTANT