Answer to Question #315665 in Calculus for Niu_bi

Question

ACTIVITY IN BASIC CALCULUS

BASIC RULES IN DERIVATIVE

 Complete the blanks of the given function below with a number (except 0 and 1) to create your own problem and find the derivative of the function. Show your complete solution to each problem.

 

1. f(x) = -₄x⁵+ ______x^-4 - 2468

2. f (x) =____x^-3- _____x1/4 - 12x

3.f(x)= ____ $\sqrt[4]{x^3} - \underset{x^6}{=} + \frac{2}{3} x^6$

4.f (x) = $\underset{x^-6}{=} -$____ x² + $\sqrt[4]{x}$

Answer 1

1.

$$f(x) = -4x^5 + 3x^{-4} - 2468\\ \begin{aligned} f'(x)= & -20 x^{4}-\frac{12}{x^{5}} \end{aligned}$$

2.

$$f (x) = 6x^{-3} -12x^{1/4} - 12x\\ \begin{aligned} f'(x) = & \frac{\mathrm{d}}{\mathrm{d} x}\left[-12 x-12 \sqrt[4]{x}+\frac{6}{x^{3}}\right] \\ = & -12 \cdot \frac{\mathrm{d}}{\mathrm{d} x}[x]-12 \cdot \frac{\mathrm{d}}{\mathrm{d} x}[\sqrt[4]{x}]+6 \cdot \frac{\mathrm{d}}{\mathrm{d} x}\left[\frac{1}{x^{3}}\right] \\ = & -12 \cdot 1-12 \cdot \frac{1}{4} x^{\frac{1}{4}-1}+6(-3) x^{-4} \\ f'(x)= & -\frac{3}{x^{\frac{3}{4}}}-\frac{18}{x^{4}}-12 \end{aligned}$$

3.

$$f(x) = 5\sqrt[4]{x^3}-\frac{3}{x^6}+\frac{2}{3}x^6\\ \begin{aligned} f'(x) =& \frac{\mathrm{d}}{\mathrm{d} x}\left[\frac{2 x^{6}}{3}+5 x^{\frac{3}{4}}-\frac{3}{x^{6}}\right] \\ = & \frac{2}{3} \cdot \frac{\mathrm{d}}{\mathrm{d} x}\left[x^{6}\right]+5 \cdot \frac{\mathrm{d}}{\mathrm{d} x}\left[x^{\frac{3}{4}}\right]-3 \cdot \frac{\mathrm{d}}{\mathrm{d} x}\left[\frac{1}{x^{6}}\right] \\ =& \frac{2 \cdot 6 x^{5}}{3}+5 \cdot \frac{3}{4} x^{\frac{3}{4}-1}-3(-6) x^{-7} \\ f'(x) = & 4 x^{5}+\frac{15}{4 \sqrt[4]{x}}+\frac{18}{x^{7}} \end{aligned}$$

4.

$$f(x) = \frac{7}{x^{-6}}-4x^2 + {x^{1/4}}\\ \begin{aligned} f'(x) = &\frac{\mathrm{d}}{\mathrm{d} x}\left[7 x^{6}-4 x^{2}+\sqrt[4]{x}\right] \\ = &7 \cdot \frac{\mathrm{d}}{\mathrm{d} x}\left[x^{6}\right]-4 \cdot \frac{\mathrm{d}}{\mathrm{d} x}\left[x^{2}\right]+\frac{\mathrm{d}}{\mathrm{d} x}[\sqrt[4]{x}] \\ =& 7 \cdot 6 x^{5}-4 \cdot 2 x+\frac{1}{4} x^{\frac{1}{4}-1} \\ f'(x)=& 42 x^{5}-8 x+\frac{1}{4 x^{\frac{3}{4}}} \end{aligned}$$