Differentiate the function y=(x^3-7x^2+4)(3x^2+14)

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have to differentiate y = (x^3-7x^2+4)(3x^2+14)

Now y = (x^3-7x^2+4)(3x^2+14)

=> y' = (x^3-7x^2+4)'(3x^2+14) + (x^3-7x^2+4)(3x^2+14)'

=> y' = (3x^2 -14x)(3x^2 + 14) + (x^3-7x^2+4)*6x

=> y' = 9x^4 + 42x^2 - 42x^3 - 196x + 6x^4 - 42x^3 + 24x

=> y' = 15x^4 - 84x^3 + 42x^2 - 172x

The required derivative is y' = 15x^4 - 84x^3 + 42x^2 - 172x

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

To differentiate the given function, we'll apply the product rule.

We'll note the brackets as:

f(x) = (x^3-7x^2+4)

g(x) = (3x^2+14)

d(f*g)/dx = (df/dx)*g + f*(dg/dx)

We'll determine (df/dx):

(df/dx) = d(x^3)/dx -7 d(x^2)/dx + d(4)/dx

(df/dx) = 3x^2 - 14x + 0

(df/dx) = 3x^2 - 14x

We'll determine (dg/dx):

(dg/dx) = d(3x^2+14)/dx

(dg/dx) = 6x

We'll substitute them in the product rule:

d(f*g)/dx = (3x^2 - 14x)*(3x^2+14) + (x^3-7x^2+4)*(6x)

We'll remove the brackets:

d(f*g)/dx = 9x^4 + 42x^2 - 42x^3 - 196x + 6x^4 - 42x^3 + 24x

We'll combine like terms:

d(f*g)/dx = 15x^4 - 84x^3 + 42x^2 - 172x

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