What was the JEE Advanced 2015 differentiability question on f, h, f(h(x)) and h(f(x))?
Only f and h(f(x)) are differentiable at x=0. The function g is differentiable on all real numbers, satisfies g(0)=0, g'(0)=0 and g'(1) is nonzero. f equals g on the nonnegative reals and equals -g on the negative reals. h is defined as e raised to |x|. The task is to decide differentiability at x=0 for f, for h, for the composition f of h, and for the composition h of f.
What does the official solution say about one-sided derivatives in this JEE Advanced 2015 question?
f is differentiable at x=0 because both one-sided derivatives equal 0. h is not differentiable at x=0 because its right derivative is 1 and left derivative is -1. f(h(x)) is not differentiable at x=0 because its right derivative is g'(1) and left derivative is -g'(1). h(f(x)) is differentiable at x=0 because both one-sided derivatives equal 0.
The official method computes left and right derivatives at x=0 for each function in turn and compares them.
For f(x), f(0)=0. For δ>0, f(−δ)=−g(−δ) and f(δ)=g(δ).
Left derivative of f at 0 is
Since g'(0)=0, this limit is 0.
Right derivative of f at 0 is
This is also g'(0), so it is 0. Therefore f is differentiable at x=0.
Now consider h(x)=e^{|x|}.
Right derivative at 0 is
This limit is 1.
Left derivative at 0 is
This limit is −1.
The one-sided derivatives are not equal, so h is not differentiable at x=0.
For f(h(x)), note that h(x)=e^{|x|}>0 for all real x. Hence the positive branch of f is used.
f(h(x))=g(e^{|x|})
At x=0, the value is g(1).
Right derivative is
Since e^δ−1 behaves like δ near 0, this limit is g'(1).
Left derivative is
This limit is −g'(1).
Given g'(1)≠0, the left and right derivatives are unequal. Therefore f(h(x)) is not differentiable at x=0.
For h(f(x)), we use h(y)=e^{|y|}. Thus
h(f(x))=e^{|f(x)|}
Since f(x)=g(x) or −g(x) depending on the sign of x, in both cases |f(x)|=|g(x)|. Hence near 0,
h(f(x))=e^{|g(x)|}
and h(f(0))=1.
Right derivative is
Write it as a product.
The first factor tends to 1, and the second factor tends to 0 because g'(0)=0. Hence the right derivative is 0.
The left derivative is also 0. Therefore h(f(x)) is differentiable at x=0.
What algebraic mistake makes students mark f(h(x)) as differentiable when it is not?
Students treat the difference quotient for f(h(x)) as a direct chain-rule product g'(h(0))·h'(0) from both sides without writing separate left and right quotients. They overlook that the left-hand approach substitutes x=-δ so the denominator becomes -δ while h(-δ)=e^δ still, producing the extra negative sign that yields -g'(1).
The calculation therefore gives g'(1) on the right and -g'(1) on the left. Since g'(1) is nonzero, the two one-sided derivatives are unequal and f(h(x)) is not differentiable at 0. The blind chain-rule step hides this sign reversal.
Why does g'(0)=0 make h(f(x)) differentiable at 0 in the JEE Advanced 2015 question?
g'(0)=0 forces the factor |g(δ)|/δ to tend to 0, which drives both one-sided limits for h(f(x)) to 0.
If g'(0) were nonzero then |g(δ)|/δ would tend to a nonzero number and the whole limit would be nonzero. With g'(0)=0 the factor |g(δ)|/δ →0 faster than the (e^u-1)/u term approaches 1, forcing both one-sided derivatives to 0.
Explicit rewriting of the right and left quotients uses |f(x)|=|g(x)| to reach the identical product form
for both sides. The second factor vanishes only because g'(0)=0.
Which two practice problems test the same one-sided derivative ideas with absolute values?
Let k(x)=x^2 sin(1/x) for x≠0, k(0)=0. Check differentiability of e^{|k(x)|} at x=0. The structure matches the h(f(x)) case exactly; k'(0)=0 forces the product limit to zero from both sides, so the composition is differentiable at zero.
Suppose p is differentiable, p(0)=0, p'(0)=0. Define q(x)=p(x) for x≥0 and q(x)=-p(x) for x<0. Is r(x)=q(e^{|x|}) differentiable at 0? Here the inner absolute value again maps to the positive branch, but the outer q introduces a potential sign flip that must be checked with separate one-sided quotients.
Numerical check: if g'(1)=3 in the original, the right derivative of f(h(x)) at 0 is 3 and the left derivative is -3.
Which later JEE questions on limit, continuity and differentiability use the same first-principles technique?
Solve the 2016 question solution next to see how one-sided limits appear with a different piecewise function: 2016 question solution. Then tackle the 2018 version: 2018 question solution. In both cases the safest route is first-principles one-sided limits at any point involving absolute value instead of blind differentiation formulas.
You can search every JEE Main paper from 2002 and every Advanced paper from 2007, by year, subject or chapter, each with a worked solution (free) inside the past-paper archive for more on this pattern. If a similar composition still feels unclear, photograph the doubt for a step-by-step reply with the necessary one-sided calculations.
Frequently asked questions
Is f differentiable at x=0 in the JEE Advanced 2015 question?
Yes. Both left and right derivatives of f at 0 equal 0. The left uses f(-δ) = -g(-δ) so the quotient is g(δ)/δ which approaches g'(0) = 0. The right quotient is also g(δ)/δ which equals 0. Hence f is differentiable at x=0.
Why is h(x) = e^|x| not differentiable at x=0?
The right derivative is lim (e^δ - 1)/δ = 1 as δ→0+. The left derivative is lim (e^{-δ} - 1)/(-δ) which simplifies to -1. Since the one-sided derivatives are unequal, h is not differentiable at x=0.
Why is f(h(x)) not differentiable at x=0 in JEE 2015?
h(x) = e^|x| > 0 so f(h(x)) = g(e^|x|). The right derivative at 0 is g'(1). The left derivative substitutes x = -δ producing an extra negative sign yielding -g'(1). Since g'(1) ≠ 0 the one-sided derivatives differ, so f(h(x)) is not differentiable at 0.
Why does g'(0)=0 make h(f(x)) differentiable at 0?
h(f(x)) = e^|f(x)| = e^|g(x)| near 0. Both one-sided derivatives reduce to the product [(e^|g(δ)| - 1)/|g(δ)|] × [|g(δ)|/δ]. The first factor → 1 while the second → 0 precisely because g'(0)=0. Thus both sides equal 0 and h(f(x)) is differentiable at 0.
What mistake do students make with f(h(x)) in the 2015 differentiability question?
Students apply the chain rule blindly from both sides and obtain g'(1)·h'(0) without separating left and right quotients. The left-hand substitution x = -δ keeps h(-δ) = e^δ but changes the denominator to -δ, producing the factor -g'(1). This sign reversal is missed, leading to the wrong conclusion that f(h(x)) is differentiable.