Find k^(')(x) if k(x)=e^((-x^(1//2)+(1)/(5)x^(-3//5)))

Question

Find  k^(')(x) if  k(x)=e^((-x^(1//2)+(1)/(5)x^(-3//5)))

Bytelearn · Accepted Answer

Identify Functions: To find the derivative of the function k(x) = e^(-x^(1/2) + x^(-3/5)), we will use the chain rule. The chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.Derivative of Outer Function: First, let's identify the outer function and the inner function. The outer function is e^u, where u is the inner function. The inner function is u(x) = -x^(1/2) + x^(-3/5).Derivative of Inner Function: The derivative of the outer function e^u with respect to u is e^u. So, (d/du)(e^u) = e^u.Derivative of -x^(1/2): Now, we need to find the derivative of the inner function u(x) = -x^(1/2) + x^(-3/5). We will differentiate each term separately using the power rule, which states that (d/dx)(x^n) = nx^(n-1).Derivative of x^(-3/5): The derivative of -x^(1/2) with respect to x is (-1/2)x^(1/2 - 1) = (-1/2)x^(-1/2).Combine Derivatives: The derivative of x^(-3/5) with respect to x is (-3/5)x^(-3/5 - 1) = (-3/5)x^(-8/5).Apply Chain Rule: Combining the derivatives of both terms, we get the derivative of the inner function u'(x) = (-1/2)x^(-1/2) - (3/5)x^(-8/5).Substitute into Expression: Now, we apply the chain rule. The derivative of k(x) with respect to x is the derivative of the outer function evaluated at the inner function times the derivative of the inner function. So, k'(x) = e^u * u'(x).Simplify if Possible: Substitute u(x) and u'(x) into the expression for k'(x). We get k'(x) = e^(-x^(1/2) + x^(-3/5)) * ((-1/2)x^(-1/2) - (3/5)x^(-8/5)).Final Answer: Simplify the expression for k'(x) if possible. However, in this case, the expression is already in its simplest form. So, the final answer is k'(x) = e^(-x^(1/2) + x^(-3/5)) * ((-1/2)x^(-1/2) - (3/5)x^(-8/5)).

Find $k^{\prime}(x)$ if $k(x)=e^{\left(-x^{1 / 2}+\frac{1}{5} x^{-3 / 5}\right)}$

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Find $k^{\prime}(x)$ if $k(x)=e^{\left(-x^{1 / 2}+\frac{1}{5} x^{-3 / 5}\right)}$