Find the derivative of f(x)=e^(sin(1))+(sin(1))^x
To find the derivative of f(x) = e^(sin(1)) + (sin(1))^x, we can use the rules of differentiation.
Let's break down the function step by step:
Step 1: Derivative of e^(sin(1))
The derivative of e^u, where u is a function of x, can be found by using the chain rule. The chain rule states that the derivative of e^u with respect to x is e^u times the derivative of u with respect to x. In this case, u = sin(1). So, the derivative of e^(sin(1)) is e^(sin(1)) times the derivative of sin(1) with respect to x.
The derivative of sin(1) with respect to x is 0, since sin(1) is a constant. Therefore, the derivative of e^(sin(1)) is e^(sin(1)) * 0 = 0.
Step 2: Derivative of (sin(1))^x
To find the derivative of (sin(1))^x, we can use the exponential rule. According to the exponential rule, the derivative of a^x, where a is a constant, is a^x times the natural logarithm of a, multiplied by the derivative of x with respect to x.
In this case, a is sin(1). So, the derivative of (sin(1))^x is (sin(1))^x times the natural logarithm of sin(1), multiplied by the derivative of x with respect to x, which is 1.
So, the derivative of (sin(1))^x is (sin(1))^x * ln(sin(1)).
Step 3: Add the derivatives from Step 1 and Step 2
Now, we can add the derivatives from Step 1 and Step 2 to find the derivative of the entire function:
Derivative of f(x) = e^(sin(1)) + (sin(1))^x
= 0 + (sin(1))^x * ln(sin(1))
Therefore, the derivative of f(x) = e^(sin(1)) + (sin(1))^x is (sin(1))^x * ln(sin(1)).