What are the first and second derivatives of #f(x)=5^((x^5)-9x)#?

Answer 1

Following the rule to differentiate exponential functions: #(d(a^b))/(dx)=a^b(ln(a))b'#, we can proceed.

We'll rename #u=x^5-9x# so that #f(x)=5^u#
#(dy)/(dx)=5^u(ln(5))u'#
Substituting #u# and #u'#:
#(dy)/(dx)=5^(x^5-9x)ln(5)(5x^4-9)#

As for the second derivative, we must see we have a three terms product. To differentiate it, we'll consider two of them as one, in a sort of chain rule, and then derivate these two as well, as shown in the formula below:

#(abc)'=(ab)'c+(ab)c'=a'bc+ab'c+abc'#
Let's just identify things here. Following the general formula: #color(red)(a=5^(x^5-9x))#, #color(blue)(b=ln(5))#, #color(green)(c=5x^4-9)#
#(dy^2)/(d^2x)=color(red)(5^(x^5-9x)ln(5)(5x^4-9))color(blue)((ln(5)))color(green)((5x^4-9))+cancel(color(red)(5^(x^5-9x))color(blue)((0))color(green)((5x^4-9)))+color(red)(5^(x^5-9x))color(blue)((ln(5)))color(green)((20x^3))#

As zero cancels the product of the second term, we have left:

#(dy^2)/(d^2x)=(5^(x^5-9x)ln(5))((5x^4-9)ln(5)+20x^3)#
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Answer 2

To find the first and second derivatives of ( f(x) = 5^{(x^5 - 9x)} ):

  1. First derivative: [ f'(x) = (\ln 5) \cdot 5^{(x^5 - 9x)} \cdot (5x^4 - 9) ]

  2. Second derivative: [ f''(x) = (\ln 5)^2 \cdot 5^{(x^5 - 9x)} \cdot ((5x^4 - 9)^2 + 20x^3) ]

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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.

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.

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.

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.

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