The derivative of with respect to is
We compute the derivative of a product.
The Product Rule
In words, the derivative of a product is the derivative of the first times the second
plus the first times the derivative of the second.
Use the product rule with and With these choices, we have and The product rule then gives
Use the product rule with and With these choices, we have and The product rule then gives
We write as a product: with To use the product rule, we need the derivatives: We can now write the derivative: \begin{align*} h'(x) &= f'(x)g(x) + f(x)g'(x) \\ &= 3x^2\sin (x) + x^3\cos (x). \end{align*}
We write as a product: with To use the product rule, we need the derivatives We can now write the derivative: \begin{align*} h'(x) &= f'(x)g(x) + f(x)g'(x)\\ &= 15x^4 \cos (x) +3x^5(-\sin (x)) \\ &= 15x^4 \cos (x) - 3x^5\sin (x). \end{align*}
We write as a product: with To use the product rule, we need the derivatives: We can now write the derivative: \begin{align*} h'(x) &= f'(x)g(x) + f(x)g'(x) \\ &= (2x - 5)e^x+ (x^2 - 5x - 2)e^x \\ &= (x^2 - 5x - 2 + 2x - 5)e^x \\ &= (x^2 - 3x - 7)e^x. \end{align*}
We write as a product: with To use the product rule, we need the derivatives: We can now write the derivative: \begin{align*} h'(x) &= f'(x)g(x) + f(x)g'(x) \\ &= 1 \cdot \ln (x) + x\frac{1}{x} \\ &= \ln (x) + 1. \end{align*}
We write as a product: with To use the product rule, we need the derivatives: We can now write the derivative: \begin{align*} h'(x) &= f'(x)g(x) + f(x)g'(x) \\ &= e^x \sin (x) + e^x \cos (x) \\ &= e^x[\sin (x) + \cos (x)]. \end{align*}
We conclude this section with a derivation of the product rule using the definition of the derivative. \begin{align*} \left [f(x)g(x)\right ]' &= \lim _{h \to 0} \frac{f(x+h)g(x+h)-f(x)g(x)}{h}\\[5pt] &= \lim _{h \to 0} \frac{f(x+h)g(x+h)- f(x)g(x+h) + f(x)g(x+h) - f(x)g(x)}{h}\\[5pt] &= \lim _{h \to 0} \frac{f(x+h)g(x+h)- f(x)g(x+h)}{h} + \frac{f(x)g(x+h) - f(x)g(x)}{h} \\[5pt] &= \lim _{h \to 0} \frac{g(x+h)\left [f(x+h)- f(x)\right ]}{h} + \frac{f(x)\left [g(x+h) - g(x)\right ]}{h}\\[5pt] &= g(x)f'(x) + f(x)g'(x). \end{align*}
2024-09-27 13:55:03