Differentiate y= 2^x

Initially this looks unlike all the other differentiation questions and seems unsolvable. However the expression 2^x can be rewritten in an equivalent form that will allow us to use the differentiation rules we already know. We know that e^(ln(x)) is the same as x, consequently e^(ln(2^x)) is 2^x. We know how to differentiate e to the power of a function of x by using the chain rule. If y=e^u, where u= ln(2^x), (this can be rewritten as 2lnx) then we have dy/du= e^u and du/dx= ln2. Multiplying these together to get dy/dx= ln2e^u. The u has to be converted back to its x form, (u=ln(2^x)), dy/dx= ln22^x. As long as the first step is remembered the rest is just applying the differentiation rules we already know.

Answered by Max G. Maths tutor

7349 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

When solving a trigonometric equation, like sin(x) = -1/3 for 0 ≤ x < 2π, why do I get an answer outside the range? Why are there many correct answers for the value of x?


Calculate the volume of revolution generated by the function, f(x) = (3^x)√x, for the domain x = [0, 1]


Solve the following equations. Leave answers in simplest terms a)e^(3x-9)=8. b) ln(2y+5)=2+ln(4-y)


Differentiate the function; f(x)=1/((5-2x^3)^2)


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences