What is the derivative of sqrt(t^5) + root(4)(t^9)√t5+4√t9? Calculus Basic Differentiation Rules Chain Rule 1 Answer marfre Jun 12, 2018 f'(t) = 5/2 t^(3/2) + 9/4 t^(5/4) or " "f'(t) = t^(5/4)/4 (10 root(4)(t) + 9) Explanation: Given: f(t) = sqrt(t^5) + root(4)(t^9) Since sqrt(x) = x^(1/2) " and " root(4)(x) = x^(1/4): f(t) = t^(5/2) + t^(9/4) f'(t) = 5/2 t^(5/2 - 2/2) + 9/4 t^(9/4 - 4/4) = 5/2 t^(3/2) + 9/4 t^(5/4) f'(t) = 10/4 t^(6/4) + 9/4 t^(5/4) f'(t) = 1/4t^(5/4) (10t^(1/4) + 9) f'(t) = t^(5/4)/4 (10 root(4)(t) + 9) Answer link Related questions What is the Chain Rule for derivatives? How do you find the derivative of y= 6cos(x^2) ? How do you find the derivative of y=6 cos(x^3+3) ? How do you find the derivative of y=e^(x^2) ? How do you find the derivative of y=ln(sin(x)) ? How do you find the derivative of y=ln(e^x+3) ? How do you find the derivative of y=tan(5x) ? How do you find the derivative of y= (4x-x^2)^10 ? How do you find the derivative of y= (x^2+3x+5)^(1/4) ? How do you find the derivative of y= ((1+x)/(1-x))^3 ? See all questions in Chain Rule Impact of this question 2071 views around the world You can reuse this answer Creative Commons License