How to find the derivative of 2 ln (x^2 -3x +4)?

Question

and.. 
1.) (5x^2-3)^6 do I use the product derivative here? 

2) Find f'(x) and the equation of the line tangent to the graph of f(x) = e^√x at x = 1

Comments

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Answer 1

The function 2 ln(x² - 3x + 4)

Differentiating with respect to x.

Apply chain rule in derivatives.

Apply the formulae d/dx[ln(x)] = 1/x

d/dx(xⁿ) =nx^n-1

d/dx (constant) = 0.

= 2 [1/(x² - 3x + 4)] d/dx(x² - 3x + 4)

= [2/(x² - 3x + 4)] (2x - 3)

= 2(2x - 3)/(x² - 3x + 4)

Derivative of 2 ln(x² - 3x + 4) = (4x - 6)/(x² - 3x + 4).

Answer 2

 1) (5x² - 3)⁶

Differentiating with respect to x.

= 6(5x² - 3)^6-1 d/dx(5x² - 3)

= 6(5x² - 3)⁵ ( 10x)

= 60x (5x² - 3)⁵

Answer 3

2) The curve f(x) = e^√x

Using the x - value find the corresponding y - value with the curve.

At x = 1, y = e^√1

y = e

The curve f(x) = e^√x

Differentiating on each side with respect to x.

f'(x) = e^√x d/dx(1/√x)

y' = e^√x (1/2√x)

y' = (e^√x)/(2√x)

Substitute the value x  = 1 in above equation.

y' = (e^√1)/(2√1)

y' = e/2

This is the slope of tangent line to the curve at  (1, e).

To find the tangent line equation, substitute the values of m = 1.35 and (x, y) = (1, e)  in the slope intercept form of an equation y = mx + b.

e = e/2(1) + b

e - e/2 = b

b = e/2

Substitute m = e/2 and b = e/2 in y = mx + b.

Tangent line is y = (e/2)x + (e/2)

By using calculator the approximate value of e/2 = 1.35.

Equation of tangent line is y = 1.35x + 1.35.