Find the derivative of the function y(x)=c^x+x^c. Assume that c is a constant.
y(x)=c^x+x^c
=y'(x)=x*c+2x
=y'(x)=cx+2x
We have c^x, not c*x
y' = c^x ln c + cx^(c-1)
= c^x (ln c + c/x)
To find the derivative of the function y(x) = c^x + x^c, we can differentiate each term separately using the rules of differentiation.
First, let's differentiate the term c^x. Recall that the derivative of a constant raised to a variable power is given by the formula (d/dx) (a^x) = (ln(a)) * (a^x), where ln(a) is the natural logarithm of a.
So, applying this rule, the derivative of c^x with respect to x is: (d/dx) (c^x) = (ln(c)) * (c^x).
Next, let's differentiate the term x^c. When the base is a variable and the exponent is a constant, we can use the power rule, which states that (d/dx) (x^n) = n * x^(n-1).
Using the power rule, the derivative of x^c with respect to x is: (d/dx) (x^c) = c * x^(c-1).
Finally, to find the derivative of the entire function y(x) = c^x + x^c, we sum up the derivatives of each term: (d/dx) (y(x)) = (d/dx) (c^x) + (d/dx) (x^c).
Substituting the derivatives we found earlier, the final derivative is: (d/dx) (y(x)) = (ln(c)) * (c^x) + c * x^(c-1).
Therefore, the derivative of the function y(x) = c^x + x^c is (ln(c)) * (c^x) + c * x^(c-1).
To find the derivative of the function y(x) = c^x + x^c, where c is a constant, we can differentiate each term separately using the rules of differentiation.
First, let's differentiate the term c^x. Since c is a constant, we can treat it as such and apply the power rule of differentiation.
The power rule states that if we have a function of the form f(x) = a^x, where a is a constant, the derivative is given by:
f'(x) = ln(a) * a^x
In our case, we have c^x, and the derivative will be:
d/dx [c^x] = ln(c) * c^x
Next, let's differentiate the term x^c. Now, we need to use the chain rule because the base is no longer a constant. The chain rule states that if we have a function of the form f(g(x)), the derivative is given by:
[f(g(x))]' = f'(g(x)) * g'(x)
In our case, f(g(x)) is x^c, and g(x) is x. Let's apply the chain rule:
[f(g(x))]' = c * x^(c-1) * 1
= c * x^(c-1)
Combining the derivatives of both terms, the derivative of y(x) = c^x + x^c is:
y'(x) = ln(c) * c^x + c * x^(c-1)
So, the derivative of the function y(x) = c^x + x^c is ln(c) * c^x + c * x^(c-1).