Home > Calculus > First Principles Example 3: square root of x

How do you derive #f(x)=x^53^(x^2-3x+2)#?

Answer 1

Ezekiel Alexander

# f'(x) = (5+(2x^2-3x)ln3)( x^4*3^(x^2-3x+2) )#

We have:

# f(x) = x^5*3^(x^2-3x+2) #

Take Natural Logarithms of both sides:

# ln f = ln {x^5*3^(x^2-3x+2)} # # " " = ln (x^5) + ln (3^(x^2-3x+2)) # # " " = (5ln x) + (x^2-3x+2)ln3 #

Differentiating Implicitly, we get:

# 1/f * (df)/(dx) = 5/x+(2x-3)ln3# # " " = (5+(2x-3)xln3)/x# # " " = (5+(2x^2-3x)ln3)/x #

And so:

# (df)/(dx) = (5+(2x^2-3x)ln3)/xf # # " = " (5+(2x^2-3x)ln3)/x x^5*3^(x^2-3x+2) # # " = " (5+(2x^2-3x)ln3)( x^4*3^(x^2-3x+2) )#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How do you derive #f(x)=x^5*3^(x^2-3*x+2)#?

How do you derive #f(x)=x^53^(x^2-3x+2)#?