Find the derivative of the following function using logarithmic differentiation?

Question

Isaiah Allen · Accepted Answer

see below

Steps in Logarithmic Differentiation 1. Take natural logarithms of both sides of an equation y=f(x) and use the Laws of Logarithms to simplify. 2. Differentiate implicitly with respect to x 3. Solve the resulting equation for y’ 4. Replace y with original equation Let #y=(5sqrtx (x^2+1)^4)/((2x+1)^3tan^5(x^2))# and lets rewrite the equation as #y=(5x^(1/2)(x^2+1)^4)/((2x+1)^3tan^5(x^2))# Now use the following Properties of Logarithms to expand the original problem #color(red)(log_b(xy)=log_b x+log_b y# #color(red)(log_b(x/y)=log_b x-log_b y# #color(red)(log_bx^n=n log_b x# That is, #color(blue)(ln y=ln((5x^(1/2)(x^2+1)^4)/((2x+1)^3tan^5(x^2)))# #color(blue)(ln y =5ln x^(1/2)+ln(x^2+1)^4-ln(2x+1)^3-ln (tan x^2)^5# #color(blue)(ln y =5/2 ln x+4 ln(x^2+1)-3ln(2x+1)-5 ln(tan x^2)# #color(blue)(d/dx (ln y =5/2 ln x+4 ln(x^2+1)-3ln(2x+1)-5 ln(tan x^2))# #color(blue)(1/y dy/dx = 5/2 1/x + 4 ((2x)/(x^2+1))-3(2/(2x+1))-5((2xsec^2(x^2))/tan x^2)# #color(blue)(1/y dy/dx = 5/(10x)+(8x)/(x^2+1)-6/(2x+1)-(10xsec^2(x^2))/tan x^2)# #color(blue)( dy/dx = [5/(10x)+(8x)/(x^2+1)-6/(2x+1)-(10xsec^2(x^2))/tan x^2]*y# #color(blue)( dy/dx = [5/(10x)+(8x)/(x^2+1)-6/(2x+1)-(10xsec^2(x^2))/tan x^2]*[(5sqrtx (x^2+1)^4)/((2x+1)^3tan^5(x^2))]#

Ezekiel Alexander · Answer

undefined To find the derivative of the given function using logarithmic differentiation, follow these steps:

1. Take the natural logarithm of both sides of the equation.
2. Use the properties of logarithms to simplify the expression.
3. Differentiate both sides with respect to x.
4. Solve for dy/dx.

The given function is:
$$ y = x^{x^2} $$

Taking the natural logarithm of both sides:
$$ \ln(y) = \ln(x^{x^2}) $$

Using the property $ \ln(a^b) = b \ln(a) $:
$$ \ln(y) = x^2 \ln(x) $$

Differentiating both sides with respect to x:
$$ \frac{1}{y} \cdot \frac{dy}{dx} = 2x \ln(x) + x^2 \cdot \frac{1}{x} $$

Simplifying the right side:
$$ \frac{1}{y} \cdot \frac{dy}{dx} = 2x \ln(x) + x $$

Multiplying both sides by y:
$$ \frac{dy}{dx} = y \left(2x \ln(x) + x\right) $$

Substituting back the original expression for y:
$$ \frac{dy}{dx} = x^{x^2} \left(2x \ln(x) + x\right) $$

So, the derivative of the function using logarithmic differentiation is:
$$ \frac{dy}{dx} = x^{x^2} \left(2x \ln(x) + x\right) $$