If your notation is what I think it is, then convert the sixth root of x^5 to exponents:
f(x) = x^(5/6)
Then use power rule
f ' (x) = (5/6)x^(-1/6)
is the x^5 under the square root? or (x^(1/2))^5?
if x^5 is under square root then the power of x is 5/2 and the derivative would be 6*5/2*x^(3/2) it actually works out the same the other way as well.
Given: f(x)= 6√x^5
it can be written as
f(x) = 6x^(5/2)
to find the derivative, first, pull the exponent (5/2) and multiply it with the coefficient (6). second, subtract 1 from the exponent.
f'(x) = [6(5/2)]x^[(5/2)-1]
= 15x(3/2)
Answer: f'(x) = 15x(3/2)
there are other ways to find the derivative, but this method is guaranteed to work every time, plus it takes less time.
f ( x ) = 6 x^(5/2)
f ` ( x ) = 15 x^(3/2)
F(x) = 6 sqrt(x^5) = 6*x^(5/2)
F'(x) = 6*(5/2)*x^((5/2)-1))
= 30/2* x^(3/2)
=15*x^(3/2)
I hope this helps.
f(x)=6(x)^5/2
f''(x)=6*(5/2) (1)(x)^(5/2-1)
f''(x)=15 x ^(3/2)
f''(x)=15√x^3
f'(x) = 15x^(3/2)
wait... Is that the 6th root, or 6 X sqrt (x^5) ???
(15 x^4)/Sqrt[x^5]
9 bro thts the answer
f'(x)=5x^(3/2)
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Verified answer
If your notation is what I think it is, then convert the sixth root of x^5 to exponents:
f(x) = x^(5/6)
Then use power rule
f ' (x) = (5/6)x^(-1/6)
is the x^5 under the square root? or (x^(1/2))^5?
if x^5 is under square root then the power of x is 5/2 and the derivative would be 6*5/2*x^(3/2) it actually works out the same the other way as well.
Given: f(x)= 6√x^5
it can be written as
f(x) = 6x^(5/2)
to find the derivative, first, pull the exponent (5/2) and multiply it with the coefficient (6). second, subtract 1 from the exponent.
f(x) = 6x^(5/2)
f'(x) = [6(5/2)]x^[(5/2)-1]
= 15x(3/2)
Answer: f'(x) = 15x(3/2)
there are other ways to find the derivative, but this method is guaranteed to work every time, plus it takes less time.
f ( x ) = 6 x^(5/2)
f ` ( x ) = 15 x^(3/2)
F(x) = 6 sqrt(x^5) = 6*x^(5/2)
F'(x) = 6*(5/2)*x^((5/2)-1))
= 30/2* x^(3/2)
=15*x^(3/2)
I hope this helps.
f(x)=6(x)^5/2
f''(x)=6*(5/2) (1)(x)^(5/2-1)
f''(x)=15 x ^(3/2)
f''(x)=15√x^3
f'(x) = 15x^(3/2)
wait... Is that the 6th root, or 6 X sqrt (x^5) ???
(15 x^4)/Sqrt[x^5]
9 bro thts the answer
f'(x)=5x^(3/2)