Answer to Question #151155 in Calculus for Angelo

Question

Differentiate the Logarithmic functions:

1. 𝑦 = 𝑥^3𝑙𝑛^2𝑥 + 𝑥^2𝑙𝑛𝑥^3 + 1

2. 𝑔(𝑥) = 𝑙𝑛 (𝑥^2+1/𝑥^3+5)

3. ℎ(𝑥) = 𝑙𝑜𝑔(𝑐𝑜𝑡4𝑥 − 𝑐𝑠𝑐4𝑥)

4. 𝑦 = 𝑙𝑛(𝑥^3𝑠𝑖𝑛2𝑥)

5. 𝑦 = 𝑙𝑛(𝑐𝑜𝑠5𝑥 + 𝑠𝑖𝑛5𝑥)

Answer 1

1.

$y'=(𝑥^3 \ln^2x + 𝑥^2\ln x^3+ 1)'$

$=3x^2 \ln^2x+x^3(2\ln x)(\dfrac{1}{x})+2x(3\ln x)+3x^2(\dfrac{1}{x})+0$

$=3x^2 \ln^2x+2x^2\ln x+6x\ln x+6, x>0$

2.

$𝑔'(𝑥) = (\ln (𝑥^2+1/𝑥^3+5))'$

$=\dfrac{1}{x^2+\dfrac{1}{x^3}+5}(2x-\dfrac{3}{x^4 }+0)$

$=\dfrac{2x^5-3}{x^6+5x^4+x}$

3.

$ℎ'(𝑥) = (\log(\cot4𝑥 − \csc4𝑥))'$

$= \dfrac{1}{\ln 10}\cdot\dfrac{4(-\dfrac{1}{\sin^24x})-4(-\dfrac{\cos4x}{\sin^24x})}{\cot4𝑥 − \csc4𝑥}$

$= \dfrac{4}{\ln 10}\cdot\dfrac{-1+\cos 4x}{\sin 4x(\cos4𝑥 −1)}$

$= \dfrac{4}{\ln 10}\csc 4x$

$\cot4𝑥 − \csc4𝑥>0=>\dfrac{\cos 4x-1}{\sin 4x}>0$

$=>\dfrac{1-\cos 4x}{\sin 4x}<0=>\sin 4x<0$

4.

$y' = (\ln(𝑥^3\sin 2𝑥))'=$

$=\dfrac{3x^2\sin 2x+2x^3\cos 2x}{x^3\sin 2x}$

$=\dfrac{3}{x}+2\cot 2x, \ x\sin 2x>0$

5.

$y' =(\ln(\cos5𝑥 + \sin5𝑥))'$

$=\dfrac{1}{\cos5𝑥 + \sin5𝑥}(-5\sin 5x+5\cos 5x)$

$=5\cdot\dfrac{\cos 5x-\sin 5x}{\cos5𝑥 + \sin5𝑥},\ \cos5𝑥 + \sin5𝑥>0$