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Find k(x)k'(x).\newlinek(x)=ex(x35)k(x)=e^{x}(-x^{\frac{3}{5}})

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Q. Find k(x)k'(x).\newlinek(x)=ex(x35)k(x)=e^{x}(-x^{\frac{3}{5}})
  1. Identify Components Requiring Product Rule: Identify the components of the function k(x)k(x) that will require the use of the product rule for differentiation.\newlineThe function k(x)=ex(x35)k(x) = e^{x}(-x^{\frac{3}{5}}) is a product of two functions: exe^{x} and (x35)(-x^{\frac{3}{5}}). The product rule states that the derivative of a product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function.
  2. Differentiate First Function: Differentiate the first function, which is exe^{x}.\newlineThe derivative of exe^{x} with respect to xx is exe^{x}.
  3. Differentiate Second Function: Differentiate the second function, which is x35-x^{\frac{3}{5}}. Using the power rule, the derivative of x35-x^{\frac{3}{5}} with respect to xx is (35)(x25)\left(\frac{3}{5}\right)(-x^{-\frac{2}{5}}).
  4. Apply Product Rule: Apply the product rule.\newlineUsing the product rule, the derivative of k(x)k(x) is:\newlinek(x)=ex(35)(x25)+ex(x35)k'(x) = e^{x} \cdot \left(\frac{3}{5}\right)(-x^{-\frac{2}{5}}) + e^{x} \cdot (-x^{\frac{3}{5}})
  5. Simplify Expression: Simplify the expression.\newlinek(x)=(35)(exx25)exx35k'(x) = \left(\frac{3}{5}\right)(-e^{x}x^{-\frac{2}{5}}) - e^{x}x^{\frac{3}{5}}\newlineThis can be further simplified by factoring out exe^{x}:\newlinek(x)=ex((35)(x25)x35)k'(x) = e^{x}\left(\left(\frac{3}{5}\right)(-x^{-\frac{2}{5}}) - x^{\frac{3}{5}}\right)
  6. Check for Errors: Check for any mathematical errors in the differentiation process. Reviewing the steps, the differentiation was done correctly using the product rule and the power rule. There are no mathematical errors.

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