If is a function defined for all and such that
prove that we must have
for some constant .
Proof. Let . Then we can compute the derivative,
Since we know we then have
Therefore, for some constant and we have
If is a function defined for all and such that
prove that we must have
for some constant .
Proof. Let . Then we can compute the derivative,
Since we know we then have
Therefore, for some constant and we have
Consider the function for a constant . First, show that and then prove
Proof. If then we have
From the definition of the derivative we also know
Define a function for by
Prove that
Proof. First, we compute and using the definition of above, and the properties of the exponential established in the previous exercise (Section 6.17, Exercise #36), we have
Therefore,
A real number raised to a real exponent is defined by
Prove the following properties:
Therefore, since we have , and so
Conversely, if then we have
Find the derivative of the function
Rewrite each of the exponentials in the expression for using the definition of the exponential function,
Now, we take the derivative directly using the chain rule and the formula for the derivative of the exponential function
Find the derivative of the following function:
We compute using the chain rule and the formulas for derivatives of logarithms and exponentials,
Find the derivative of the following function:
We can compute this derivative directly using the quotient rule,
Find the derivative of the following function:
We can compute this derivative by multiplying out the expression for and using the product and chain rules. First, we multiply out the expression,
Then, we take the derivative,
Let and be constants with at least one of them nonzero and define
Using integration by parts, establish the following formulas for constants ,
Using these formulas prove the following integration formulas,
To establish the formula we use integration by parts letting
Then we can evaluate using the formula for integration by parts,
To establish the second formula, , we use integration by parts again. Let
Then we have
This establishes the two requested equations, now we prove the two integral identities.
Proof. Solving for in the second equation above we have
Plugging this into the first equation we have
Next, for the second integral equation we are asked to prove, we use the formula we obtained for above,
Then, we use the expression we obtained for into this,
This implies,
Find values for the constants and such that the following equation is true:
To find values of the constants we evaluate the integral,
Therefore, we need to find constants such that
Therefore, we can choose any value for and then set and this pair will make the equation true for all .