Question 6.7.3: Find h'(x) when h(x) = (x³ − x) · (5x^4 + x²). Confirm the a......
Find h^{\prime}(x) when h(x) = (x^{3} − x) · (5x^{4} + x^{2}). Confirm the answer by expanding h(x) as a single polynomial, then differentiating the result.
We see that h(x) = f (x) · g(x) with f (x) = x³ − x and g(x) = 5x^{4} + x^{2}. Here f^{\prime}(x) = 3x^{2} − 1 and g^{\prime}(x) = 20x^{3} + 2x. Thus, from (6.7.2)
F(x)=f(x)\cdot g(x)\Rightarrow F^{\prime}(x)=f^{\prime}(x)\cdot g(x)+f(x)\cdot g^{\prime}(x) (6.7.2)
h^{\prime}(x)=f^{\prime}(x)\cdot g(x)+f(x)\cdot g^{\prime}(x)=(3x^{2}-1)\cdot(5x^{4}+x^{2})+(x^{3}-x)\cdot(20x^{3}+2x)Usually we simplify the answer by expanding to obtain just one polynomial. Simple computations give h^{\prime}(x) = 35x^{6} − 20x^{4} − 3x^{2}.
Alternatively, expanding h(x) as a single polynomial gives h(x) = 5x^{7} − 4x^{5} − x^{3}, whose derivative, from rules (6.6.4) and (6.7.1), is h^{\prime}(x) = 35x^{6} − 20x^{4} − 3x^{2}.
f(x)=x^{a}\Rightarrow f^{\prime}(x)=a x^{a-1} (6.6.4)
F(x)=f(x)\pm g(x)\Rightarrow F^{\prime}(x)=f^{\prime}(x)\pm g^{\prime}(x) (6.7.1)