Answer to Question #165699 in Calculus for Robert John Hermoso

Question

Differentiate the function or find 𝑑𝑦 𝑑𝑥 and simplify the result.

1. 𝑓(𝑥) = ln(4 + 5𝑥)

2. 𝐹(𝑦) = ln(sin 5𝑦)

3. 𝑦 = ln|𝑥³ + 1|

4. 𝑦 = 𝑥⁵(𝑥+2)/𝑥−3 (by logarithmic differentiation)

5. 𝑦 = 𝑒^5x

6. 𝑦 = 𝑒^x sin 𝑒^x

7. 𝑓(𝑥) = 3^5x

8. 𝑓(𝑥) = 4^sin2x

9. 𝑓(𝑥) = sinh 𝑥²

10. 𝑓(𝑥) = coth 1/x

Answer 1

1. $\bold{f(x) = ln(4+5x)}$

Let

$$u = 4+5x\\$$

Then

$$f(x) = ln(u)$$

By chain rule,

$$\dfrac{d}{dx}[f(x)] = \dfrac{d}{du}[f(x)] \times \dfrac{d}{dx}[u]$$

$$\dfrac{d}{du}[f(x)] = \dfrac{1}{u} = \dfrac{1}{4+5x}$$ $$\dfrac{d}{dx}[u] = 5$$ $$\therefore \bold{\dfrac{d}{dx}[f(x)]} = \dfrac{1}{4+5x} \times 5 = \bold{\dfrac{5}{4+5x}}$$

$\bold{2. \quad F(y) = \ln(\sin(5y))}$

$$\dfrac{d}{dy}[\ln(\sin(5y))]\\ = \dfrac{1}{\sin(5y)} \times \dfrac{d}{dy}[5y]\\ = \dfrac{\cos(5y) \times \frac{d}{dy}[5y]}{\sin(5y)}\\ = \dfrac{\cos(5y) \cdot 5 \cdot \frac{d}{dy}[5]}{\sin(5y)}\\ = \dfrac{5\cos(5y) \cdot 1}{\sin(5y)}\\ \bold{F'(y) = \dfrac{5\cos(5y)}{\sin(5y)}}$$

$\bold{3. \quad 𝑦 = \ln|𝑥^3 + 1|}$

$$y' = \dfrac{1}{|x^3+1|} \cdot \dfrac{d}{dx}[|x^3+1|]\\ = \dfrac{\frac{x^3+1}{|x^3+1|} \cdot \frac{d}{dx}[x^3+1]}{|x^3+1|}\\ = \dfrac{\frac{d}{dx}[x^3]+\frac{d}{dx}[1]}{x^3+1}\\ = \dfrac{3x^2+0}{x^3+1}\\ \bold{y'=\dfrac{3x^2}{x^3+1}}$$