Derivative Rules - Multi Chain Quotient General Exponential (with Rule) to Derivative

Question

Find the derivative f'(x) using the quotient, chain, and general exponential rules.

$$\begin{array}{l} \displaystyle \text{if } y = f(g(x)),\ y' = f'(g(x))\cdot g'(x) \[12pt] \displaystyle f(x) = \frac{\frac{1}{(2x^{2} - 5)}}{2 \cdot 7^{x}} \end{array}$$

Accepted Answer

$$f'(x) = \frac{-(2x^{2} - 5)^{-2}(4x)\left(2 \cdot 7^{x}\right) - \frac{1}{(2x^{2} - 5)}\left(2 \cdot 7^{x}\ln{7}\right)}{\left(2 \cdot 7^{x}\right)^{2}}$$

Answer

$$f'(x) = \frac{-(2x^{2} - 5)^{-2}(4x)\left(2 \cdot 7^{x}\right) + \frac{1}{(2x^{2} - 5)}\left(2 \cdot 7^{x}\ln{7}\right)}{\left(2 \cdot 7^{x}\right)^{2}}$$

Answer

$$f'(x) = \frac{\frac{1}{(2x^{2} - 5)}\left(2 \cdot 7^{x}\ln{7}\right) - -(2x^{2} - 5)^{-2}(4x)\left(2 \cdot 7^{x}\right)}{\left(2 \cdot 7^{x}\right)^{2}}$$

Answer

$$f'(x) = \frac{-(2x^{2} - 5)^{-2}(4x)\left(2 \cdot 7^{x}\right) - \frac{1}{(2x^{2} - 5)}\left(2 \cdot 7^{x}\ln{7}\right)}{2 \cdot 7^{x}}$$

Derivative Rules - Multi Chain Quotient General Exponential (with Rule) to Derivative

Practice:

Derivative Rules - Multi Chain Quotient General Exponential (with Rule) to Derivative