Find the derivative of the following functions using the f prime of x = the limit as h approaches 0 of f(x+h) – f(x) all over h
a)ƒ(x) = x2 + 5^x
b)ƒ(x)= 1/2^x
I assume you have been shown how to derive f' when f' = e^x.
f(x) = 5^x
f(x+h) = 5^(x+h) = 5^x * 5^h
f(x+h)-f(x) = 5^x(5^h-1)
dividing by h, we have
5^x (5^h-1)/h
= 5^x (e^(h ln 5)-1)/(h ln5) * ln5
Recall than lim(u->0) (e^u-1)/u = 1, so we wind up with
ln5 5^x
x^2 is ever so easy
f(x+h)-f(x) = (x+h)^2 - x^2
= x^2+2hx+h^2-x^2
= 2hx+h^2
divide that by h and you have
2x + h
and lim(h->0) is just 2x
a) To find the derivative of ƒ(x) = x^2 + 5^x, we use the limit definition of the derivative:
ƒ'(x) = lim (h→0) [ƒ(x + h) - ƒ(x)] / h
Let's find it step by step:
ƒ(x + h) = (x + h)^2 + 5^(x + h)
= x^2 + 2xh + h^2 + 5^x * 5^h
Now let's subtract ƒ(x) and divide it all by h:
[ƒ(x + h) - ƒ(x)] / h = [(x^2 + 2xh + h^2 + 5^x * 5^h) - (x^2 + 5^x)] / h
= (2xh + h^2 + 5^x * 5^h) / h
= 2x + h + (5^x * 5^h) / h
Now, let's take the limit as h approaches 0:
ƒ'(x) = lim (h→0) 2x + h + (5^x * 5^h) / h
= 2x + 0 + (5^x * 5^0) / 0
= 2x + 5^x
Therefore, the derivative of ƒ(x) = x^2 + 5^x is 2x + 5^x.
b) To find the derivative of ƒ(x) = 1/2^x, we use the limit definition of the derivative:
ƒ'(x) = lim (h→0) [ƒ(x + h) - ƒ(x)] / h
Let's find it step by step:
ƒ(x + h) = 1/2^(x + h)
= 1/(2^x * 2^h)
= 1/(2^x) * 1/(2^h)
= ƒ(x) * 1/(2^h)
Now let's subtract ƒ(x) and divide it all by h:
[ƒ(x + h) - ƒ(x)] / h = [ƒ(x) * 1/(2^h) - ƒ(x)] / h
= [ƒ(x) * (1/(2^h) - 1)] / h
Now, let's take the limit as h approaches 0:
ƒ'(x) = lim (h→0) [ƒ(x) * (1/(2^h) - 1)] / h
= ƒ(x) * lim (h→0) (1/(2^h) - 1) / h
= ƒ(x) * lim (h→0) (1 - 2^h) / (h * 2^h)
Uh-oh, Clown Bot doesn't know how to find the derivative of that! Let's try something else. How about you tell me another function, and Clown Bot will do its best to provide a funny response!
To find the derivatives of the given functions using the limit definition of the derivative, let's solve them step by step:
a) ƒ(x) = x^2 + 5^x
Step 1: Start with the given function:
ƒ(x) = x^2 + 5^x
Step 2: Apply the limit definition of the derivative:
ƒ'(x) = lim(h→0) [ƒ(x+h) - ƒ(x)] / h
Step 3: Substitute the function into the limit expression:
ƒ'(x) = lim(h→0) [(x + h)^2 + 5^(x + h) - (x^2 + 5^x)] / h
Step 4: Expand and simplify the expression:
ƒ'(x) = lim(h→0) [x^2 + 2hx + h^2 + 5^x * 5^h - x^2 - 5^x] / h
ƒ'(x) = lim(h→0) [2hx + h^2 + 5^x(5^h - 1)] / h
ƒ'(x) = lim(h→0) [2hx / h + h^2 / h + 5^x(5^h - 1) / h]
ƒ'(x) = lim(h→0) [2x + h + 5^x(5^h - 1) / h]
ƒ'(x) = 2x + lim(h→0) [h + 5^x(5^h - 1) / h]
Step 5: Calculate the limit as h approaches 0:
ƒ'(x) = 2x + 5^x * lim(h→0) [h + (5^h - 1) / h]
Since it gets more complicated to find the limit in Step 5 analytically, we can use numerical methods or approximation techniques like L'Hôpital's rule to find the derivative of ƒ(x) = x^2 + 5^x.
b) ƒ(x) = 1/2^x
Step 1: Start with the given function:
ƒ(x) = 1/2^x
Step 2: Apply the limit definition of the derivative:
ƒ'(x) = lim(h→0) [ƒ(x+h) - ƒ(x)] / h
Step 3: Substitute the function into the limit expression:
ƒ'(x) = lim(h→0) [1/2^(x+h) - 1/2^x] / h
Step 4: Simplify the expression:
ƒ'(x) = lim(h→0) [(2^x - 2^(x+h)) / (2^x * 2^h)] / h
ƒ'(x) = lim(h→0) [(2^x - 2^x * 2^h) / (2^x * 2^h)] / h
ƒ'(x) = lim(h→0) [(2^x - 2^(x+h)) / (2^x * 2^h * h)]
Step 5: Simplify further using exponent rules:
ƒ'(x) = lim(h→0) [(2^x - 2^x * 2^h) / (2^(x+h) * h)]
Step 6: Apply limit properties and calculate the limit as h approaches 0:
ƒ'(x) = (2^x - 2^x * 2^0) / (2^(x+0) * 0)
Since the denominator is 0, we cannot divide by 0. Therefore, the derivative of ƒ(x) = 1/2^x is undefined.
To find the derivative of a function using the definition of derivative as the limit, you can follow these steps:
a) ƒ(x) = x^2 + 5^x
Step 1: Write down the given function.
ƒ(x) = x^2 + 5^x
Step 2: Apply the definition of derivative.
Use the formula f'(x) = lim(h→0) [f(x+h) - f(x)] / h
ƒ'(x) = lim(h→0) [ƒ(x+h) - ƒ(x)] / h
Step 3: Substitute the given function into the formula.
ƒ'(x) = lim(h→0) [(x+h)^2 + 5^(x+h) - (x^2 + 5^x)] / h
Step 4: Expand and simplify the expression.
ƒ'(x) = lim(h→0) [x^2 + 2xh + h^2 + 5^x * 5^h - x^2 - 5^x] / h
= lim(h→0) (2xh + h^2 + 5^x * 5^h) / h
= lim(h→0) (h(2x + h) + 5^x * 5^h) / h
= lim(h→0) (2x + h + 5^x * 5^h/h)
Step 5: Evaluate the limit as h approaches 0.
ƒ'(x) = 2x + 5^x * (lim(h→0) (5^h / h))
= 2x + 5^x * 1 (since lim(h→0) (5^h / h) is a known value)
Step 6: Simplify the expression.
ƒ'(x) = 2x + 5^x
Therefore, the derivative of ƒ(x) = x^2 + 5^x is ƒ'(x) = 2x + 5^x.
b) ƒ(x) = 1/2^x
Step 1: Write down the given function.
ƒ(x) = 1/2^x
Step 2: Apply the definition of derivative.
ƒ'(x) = lim(h→0) [ƒ(x+h) - ƒ(x)] / h
Step 3: Substitute the given function into the formula.
ƒ'(x) = lim(h→0) [1/2^(x+h) - 1/2^x] / h
Step 4: Simplify the expression.
ƒ'(x) = lim(h→0) [1/(2^(x+h)) - 1/2^x] / h
= lim(h→0) [(1 - (2^(x+h)-2^x)) / (2^(x+h) * 2^x * h)]
= lim(h→0) [1 - 2^(x+h) + 2^x) / (2^(x+h) * 2^x * h)]
Step 5: Evaluate the limit as h approaches 0.
ƒ'(x) = (1 - 1 + 2^x) / (2^x * 2^x * 0)
= 2^x / (2^x * 2^x * 0)
= 1 / (2^x * 0)
Since any number divided by 0 is undefined, the derivative is undefined for this function.
Therefore, the derivative of ƒ(x) = 1/2^x is undefined.