Answer to Question #165699 in Calculus for Robert John Hermoso

Question #165699

Differentiate the function or find 𝑑𝑦 𝑑π‘₯ and simplify the result.

1. 𝑓(π‘₯) = ln(4 + 5π‘₯)

2. 𝐹(𝑦) = ln(sin 5𝑦)

3. 𝑦 = ln|π‘₯3 + 1|

4. 𝑦 = π‘₯5(π‘₯+2)/π‘₯βˆ’3 (by logarithmic differentiation)

5. 𝑦 = 𝑒5x

6. 𝑦 = 𝑒x sin 𝑒x

7. 𝑓(π‘₯) = 35x

8. 𝑓(π‘₯) = 4sin2x

9. 𝑓(π‘₯) = sinh π‘₯2

10. 𝑓(π‘₯) = coth 1/x


1
Expert's answer
2021-02-24T06:02:12-0500
  1. f(x)=ln(4+5x)\bold{f(x) = ln(4+5x)}

Let


u=4+5xu = 4+5x\\

Then


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

By chain rule,


ddx[f(x)]=ddu[f(x)]Γ—ddx[u]\dfrac{d}{dx}[f(x)] = \dfrac{d}{du}[f(x)] \times \dfrac{d}{dx}[u]

ddu[f(x)]=1u=14+5x\dfrac{d}{du}[f(x)] = \dfrac{1}{u} = \dfrac{1}{4+5x}ddx[u]=5\dfrac{d}{dx}[u] = 5∴ddx[f(x)]=14+5xΓ—5=54+5x\therefore \bold{\dfrac{d}{dx}[f(x)]} = \dfrac{1}{4+5x} \times 5 = \bold{\dfrac{5}{4+5x}}

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


ddy[ln⁑(sin⁑(5y))]=1sin⁑(5y)Γ—ddy[5y]=cos⁑(5y)Γ—ddy[5y]sin⁑(5y)=cos⁑(5y)β‹…5β‹…ddy[5]sin⁑(5y)=5cos⁑(5y)β‹…1sin⁑(5y)Fβ€²(y)=5cos⁑(5y)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)}}

3.𝑦=ln⁑∣π‘₯3+1∣\bold{3. \quad 𝑦 = \ln|π‘₯^3 + 1|}


yβ€²=1∣x3+1βˆ£β‹…ddx[∣x3+1∣]=x3+1∣x3+1βˆ£β‹…ddx[x3+1]∣x3+1∣=ddx[x3]+ddx[1]x3+1=3x2+0x3+1yβ€²=3x2x3+1y' = \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}}






























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