H(x) = (x^4 - 2x +7)(x^-3 + 2x^-4)
H'(x)=
The derivative of a product.
The rule is:
the first times the derivative of the second PLUS the second times the derivative of the first.
simplify your answer.
The derivative of the first: 4x-2
The derivative of the second
-3x^-4 -8x^-5
Can you finish with these hints?
To find the derivative of the function H(x), we can use the product rule, which states that if we have two functions f(x) and g(x), then the derivative of their product f(x)g(x) is given by:
(f(x)g(x))' = f'(x)g(x) + f(x)g'(x)
Let's find the derivative of each term separately:
For the first term (x^4 - 2x + 7):
- The derivative of x^4 with respect to x is 4x^(4-1) = 4x^3.
- The derivative of -2x with respect to x is -2.
- The derivative of 7 with respect to x is 0, as 7 is a constant.
So, the first term becomes: 4x^3 - 2.
For the second term (x^-3 + 2x^-4):
- To find the derivative of x^-3, we can rewrite it as 1/x^3. Then, using the power rule, the derivative is -3x^(-3-1) = -3/x^4.
- Similarly, for 2x^-4, we can rewrite it as 2/x^4, and the derivative is -4(2/x^4) = -8/x^5.
So, the second term becomes: -3/x^4 + (-8/x^5) = -3/x^4 - 8/x^5.
Now, applying the product rule, we have:
H'(x) = (4x^3 - 2)(x^-3 + 2x^-4)' + (x^4 - 2x + 7)(-3/x^4 - 8/x^5)'.
Simplifying further:
H'(x) = (4x^3 - 2)(-3/x^4 - 8/x^5) + (x^4 - 2x + 7)(12/x^5 + 40/x^6).
Finally, you can further simplify and combine like terms if necessary.