Question For j(x)=(4x^(2)-8)(-5x^(2)-8x), find j^(')(x) by applying the product rule.

Question

Question For  j(x)=(4x^(2)-8)(-5x^(2)-8x), find  j^(')(x) by applying the product rule.

Bytelearn · Accepted Answer

Identify Functions: To find the derivative of the function j(x) = (4x^2 - 8)(-5x^2 - 8x), we will apply the product rule. The product rule states that if we have two functions u(x) and v(x), then the derivative of their product u(x)v(x) is given by u'(x)v(x) + u(x)v'(x). Let's identify u(x) and v(x) in our function. u(x) = 4x^2 - 8 v(x) = -5x^2 - 8xFind Derivatives: Next, we need to find the derivatives of u(x) and v(x) with respect to x. For u(x) = 4x^2 - 8, the derivative u'(x) is found by applying the power rule, which states that the derivative of x^n is n*x^(n-1). u'(x) = d/dx [4x^2] - d/dx [8] u'(x) = 8x - 0 u'(x) = 8xApply Product Rule: Now, let's find the derivative of v(x) = -5x^2 - 8x. Again, we apply the power rule to each term. v'(x) = d/dx [-5x^2] - d/dx [8x] v'(x) = -10x - 8Expand Expression: With both derivatives u'(x) and v'(x) calculated, we can now apply the product rule to find j'(x). j'(x) = u'(x)v(x) + u(x)v'(x) Substitute the expressions we found for u'(x), v(x), u(x), and v'(x) into the formula. j'(x) = (8x)(-5x^2 - 8x) + (4x^2 - 8)(-10x - 8)Combine Like Terms: Now we will expand the terms in the expression for j'(x). j'(x) = -40x^3 - 64x^2 - 40x^3 - 32x^2 + 80x + 64Combine like terms in the expression for j'(x). j'(x) = -40x^3 - 40x^3 - 64x^2 - 32x^2 + 80x + 64 j'(x) = -80x^3 - 96x^2 + 80x + 64

Question $\newline$ For $j(x)=\left(4 x^{2}-8\right)\left(-5 x^{2}-8 x\right)$ , find $j^{\prime}(x)$ by applying the product rule.

Full solution

More problems from Multiply and divide rational expressions

Related topics

Question\newlineFor j(x)=(4x2−8)(−5x2−8x) j(x)=\left(4 x^{2}-8\right)\left(-5 x^{2}-8 x\right) j(x)=(4x2−8)(−5x2−8x), find j′(x) j^{\prime}(x) j′(x) by applying the product rule.

Question $\newline$ For $j(x)=\left(4 x^{2}-8\right)\left(-5 x^{2}-8 x\right)$ , find $j^{\prime}(x)$ by applying the product rule.