Class 12 Maths, Worked in Full Past papersGet updates
Chapter 5 · Continuity and Differentiability

Exercise 5.4 · Q6: Differentiate ex+ex2+⋯+ex5e^x+e^{x^2}+\dots+e^{x^5}

Exercise 5.4Sum rule and chain rule with eˣ · Section 5.4
The question

Differentiate the following w.r.t. xx⁠:

ex+ex2+⋯+ex5e^x + e^{x^2} + \dots + e^{x^5}
Exam answer

Let y=ex+ex2+ex3+ex4+ex5y = e^x + e^{x^2} + e^{x^3} + e^{x^4} + e^{x^5}⁠. By the chain rule, ddxexk=exk⋅kxk−1\frac{d}{dx}e^{x^k} = e^{x^k}\cdot kx^{k - 1}⁠. So

∴ dydx=ex+2xex2+3x2ex3+4x3ex4+5x4ex5\begin{aligned} \therefore\ \dfrac{dy}{dx} &= e^x + 2xe^{x^2} + 3x^2e^{x^3} \\ &\quad + 4x^3e^{x^4} + 5x^4e^{x^5} \end{aligned}

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

The function is a sum of five terms that all have the same shape, exke^{x^k} with k=1,2,3,4,5k = 1, 2, 3, 4, 5⁠. The sum rule lets us differentiate term by term, and one chain-rule calculation for a general kk handles all five terms at once: ddxexk=kxk−1exk\frac{d}{dx}e^{x^k} = kx^{k-1}e^{x^k}⁠.

Given, to find, and the plan
Given
y=ex+ex2+ex3+ex4+ex5y = e^x + e^{x^2} + e^{x^3} + e^{x^4} + e^{x^5} (the dots stand for the two missing terms ex3e^{x^3} and ex4e^{x^4}⁠).
To find
dydx\dfrac{dy}{dx}⁠, and where it is valid.
Plan
(1) Domain. (2) Sum rule: differentiate each term. (3) Chain rule for the general term exke^{x^k}⁠. (4) Apply it for k=1,…,5k = 1, \dots, 5 and add.

Basics you need

1Sum rule

(u+v)′=u′+v′(u + v)' = u' + v' (rule (1), Section 5.3); applied repeatedly, it works for any finite number of terms. Basics: derivative rules

2Theorem 5 (1) and the power rule

ddtet=et\dfrac{d}{dt}e^t = e^t (Section 5.4) and ddxxk=kxk−1\dfrac{d}{dx}x^k = kx^{k-1} (Table 5.3). Basics: exponentials and logarithms

3Chain rule

If y=f(t)y = f(t) and t=g(x)t = g(x)⁠, then dydx=dydt⋅dtdx\dfrac{dy}{dx} = \dfrac{dy}{dt}\cdot\dfrac{dt}{dx} (Theorem 4). Basics: composite functions

Solution

  1. Where is the function defined?

    Each power xkx^k is defined for every real xx⁠, and ete^t is defined for every real tt⁠. So every term, and hence yy⁠, is defined (and, as we will see, differentiable) for all x∈Rx \in \R⁠.

    No division, root or logarithm appears, so nothing has to be excluded. The dots follow the pattern set by the first two terms and the last one: the power of xx in the exponent goes up by one each time, from 11 to 55⁠.
    Domain of the exponential function is R\R (Section 5.4). Basics: exponentials and logarithms

  2. Use the sum rule

    dydx=ddxex+ddxex2+ddxex3+ddxex4+ddxex5.\begin{aligned} \frac{dy}{dx} &= \frac{d}{dx}e^x + \frac{d}{dx}e^{x^2} + \frac{d}{dx}e^{x^3} \\ &\quad + \frac{d}{dx}e^{x^4} + \frac{d}{dx}e^{x^5}. \end{aligned}
    The derivative of a sum is the sum of the derivatives, provided each term is differentiable (Step 3 shows they are). This splits one big problem into five small ones of the same type.
    Sum rule, rule (1) of Section 5.3. Basics: derivative rules

  3. One chain-rule calculation for the general term exke^{x^k}

    Fix a positive integer kk⁠. Inner: t=xkt = x^k⁠. Outer: ete^t⁠. Then

    ddxexk=et⋅dtdx=exk⋅kxk−1=kxk−1exk.\begin{aligned} \frac{d}{dx}e^{x^k} &= e^t\cdot\frac{dt}{dx} = e^{x^k}\cdot kx^{k-1} \\ &= kx^{k-1}e^{x^k}. \end{aligned}

    This holds for every real xx⁠.

    The exponential copies itself (Theorem 5), evaluated at the inner value xkx^k⁠; the chain rule then multiplies by the inner derivative kxk−1kx^{k-1} (power rule). Doing it once for a general kk saves repeating the same argument five times. For k=1k = 1 it gives 1⋅x0ex=ex1\cdot x^0e^{x} = e^x⁠, as it should.
    Chain rule, Theorem 4, Section 5.3.1; Theorem 5 (1), Section 5.4. Lesson: the chain rule

  4. Apply it for k=1,2,3,4,5k = 1, 2, 3, 4, 5 and add

    kktermderivative kxk−1exkkx^{k-1}e^{x^k}
    1exe^xexe^x
    2ex2e^{x^2}2xex22xe^{x^2}
    3ex3e^{x^3}3x2ex33x^2e^{x^3}
    4ex4e^{x^4}4x3ex44x^3e^{x^4}
    5ex5e^{x^5}5x4ex55x^4e^{x^5}
    dydx=ex+2xex2+3x2ex3+4x3ex4+5x4ex5.\begin{aligned} \frac{dy}{dx} &= e^x + 2xe^{x^2} + 3x^2e^{x^3} \\ &\quad + 4x^3e^{x^4} + 5x^4e^{x^5}. \end{aligned}
    Adding the five derivatives is exactly what the sum rule of Step 2 asked for. Writing them in a table keeps the coefficient kk and the power xk−1x^{k-1} lined up with the right term.
    Sum rule, Section 5.3. Basics: derivative rules
Result dydx=ex+2xex2+3x2ex3+4x3ex4+5x4ex5,\begin{aligned} \frac{dy}{dx} &= e^x + 2xe^{x^2} + 3x^2e^{x^3} \\ &\quad + 4x^3e^{x^4} + 5x^4e^{x^5}, \end{aligned}

valid for every real xx⁠. In short, dydx=∑k=15kxk−1exk\dfrac{dy}{dx} = \sum_{k=1}^{5} kx^{k-1}e^{x^k}⁠. (Textbook answer: ex+2xex2+3x2ex3e^x + 2xe^{x^2} + 3x^2e^{x^3} + 4x3ex4+5x4ex5+\, 4x^3e^{x^4} + 5x^4e^{x^5}⁠. The same.)

Check it with numbers

Take x=0.5x = 0.5 (at x=1x = 1 every factor xk−1x^{k-1} equals 11⁠, which would hide a mistake in the powers). The five terms of the derivative are about

1.6487213+1.2840254+0.8498613+0.5322472+0.3224198≈4.6372751.\begin{aligned} &1.6487213 + 1.2840254 + 0.8498613 \\ &\quad + 0.5322472 + 0.3224198 \approx 4.6372751. \end{aligned}

A difference quotient of yy with h=0.0001h = 0.0001 on each side of 0.50.5 gives 4.637284.63728⁠. They agree.

Common mistakes
  1. Copying the terms unchanged. Only exe^x is its own derivative; ex2e^{x^2} has derivative 2xex22xe^{x^2}⁠, not ex2e^{x^2}⁠.
  2. Dropping the power of xx⁠. ddxex4=4x3ex4\frac{d}{dx}e^{x^4} = 4x^3e^{x^4}⁠, not 4ex44e^{x^4}⁠.
  3. Changing the exponent. The exponential keeps its exponent xkx^k⁠; it is not lowered to xk−1x^{k-1}⁠.
What to take away
  1. When many terms share one shape, differentiate the general term once and then substitute.
  2. ddxexk=kxk−1exk\dfrac{d}{dx}e^{x^k} = kx^{k-1}e^{x^k}⁠: copy the exponential, multiply by the derivative of its exponent.

Every fact used, and where it comes from

FactUsed inSource
Exponential function has domain R\RStep 1Feature (1) of exponential functions, Section 5.4
Sum ruleSteps 2, 4Rule (1), Section 5.3
ddtet=et\frac{d}{dt}e^t = e^tStep 3Theorem 5 (1), Section 5.4 (Lesson: exponential and logarithmic functions)
ddxxk=kxk−1\frac{d}{dx}x^k = kx^{k-1}Step 3Table 5.3, Section 5.3
Chain ruleStep 3Theorem 4, Section 5.3.1
Found an error, or want more?

Tell us if something on this page is wrong or unclear, or if you would like a fuller explanation, another method or a new feature. Send feedback on this problem →