y= x^5+ 2x^3+ 6. If dy/dx=0 , find x.

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giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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To calculate the roots of the equation dy/dx=0, we'll have to differentiate the given relation both sides.

y= x^5+ 2x^3+ 6

dy= (x^5+ 2x^3+ 6)'dx

To differentiate the given expression x^5+ 2x^3+ 6, we'll differentiate each term of this expression, with respect to x.

(x^5+ 2x^3+ 6)' = (x^5)'+ (2x^3)'+ (6)'

To calculate the derivative of the power function;

f(x) = x^n

f'(x) = (x^n)'

But (x^n) = x*x*x*......*x, n times

(x^n)' = (x*x*x*......*x)' = x'*x*...*x + x*x'*x*...*x + ...+x*x*...*x' = n*x^(n-1)

If n = 5 => (x^5)' = 5x^4

For n = 3 => (2x^3)' = 6x^2

(6)' = 0

(x^5+ 2x^3+ 6)' = 5x^4 + 6x^2

Now, we'll calculate dy/dx = 0 <=> 5x^4 + 6x^2 = 0

We'll factorize by x^2 and we'll get:

x^2(5x^2 + 6)= 0

x^2 = 0

x1=x2 = 0

5x^2 + 6 = 0

We'll subtract 6 both sides:

5x^2 = -6 impossible!

x^2>0, for any value of x

5x^2>0 also, so it's imposible for 5x^2 to have negative values, for any x.

The real solutions of dy/dx = 0 are x1=x2 = 0.

krishna-agrawala's profile pic

krishna-agrawala | College Teacher | (Level 3) Valedictorian

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Given:

y = x^5 + 2x^3 + 6

Therefore:

dy/dx = 5x^4 + 2*3x^2 = 0 (given)

==> 5x^4 + 6x^2 = 0

==> x^2(5x^2 + 6) = 0

Therefor x^2 has following 2 possible values:

x^2 = 0 and -6/5

When x^2 = 0

x = 0

When x^2 = -6/5, x cannot be a real number. The value of x in this case is:

x = i(5/6)^1/2

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

y = x^5+2x^3+6.

To find dy/dx.

Soltion:

y= x^5+2x^3+6.

Differentiating both sides we get:

dy/dx = d/dx(x^5+2x^3+6)

dy/dx = d/dx(x^5) +d/dx(2x^3) + d/dx(6)

dy/dx = 5x^(5-1)+2 *3*x^(2-1) +0, as d/dx{k(x^n)} = k {d/dx(x^n)} = k *n*x^(n-1).

dy/dx =5x^4+6x^2

william1941's profile pic

william1941 | College Teacher | (Level 3) Valedictorian

Posted on

In this question, y or f(x) = x^5+ 2x^3+6 .Now we have to find the differential of f(x).

Here we will use a general relation which is valid whenever we are trying to differentiate a function. It states that if f(x) = a*x^n, then f’(x) is equal to a*n*x^(n-1) .

=> f'(x)= 5x^4 + 3*2*x^2 = 5x^4 + 6x^2

As this is equal to zero

=> 5x^4 + 6x^2 = 0

=> x^2 ( 5x^2 + 6)=0

Now x^2=0 => x=0

and x^2 = - 6/5 is invalid as a square is always positive.

Therefore the only valid value of x is 0.

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