Q.For the function f(x)=x3+3x2+1 given above, the tangent line at point x=−3 is drawn using GeoGebra graphing calculator. Draw the tangent line using this application at x=−2 and x=1.
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Concept understanding — Derivatives of Algebraic Functions Using Chain Rule
Derivatives of Algebraic Functions Using the Chain Rule
The Intuition: Why Do We Need This?
You already know how to differentiate simple functions like x2 or sinx. But what about something like (3x2+5)7? You could expand it — but that would be a nightmare. Or what about 1−x2? Expanding is impossible.
The chain rule solves this by letting you differentiate functions inside other functions — compositions. Think of it like peeling an onion: you differentiate the outer layer, then multiply by the derivative of the inner layer.
The Core Idea: A Real-World Analogy
Imagine a machine that first doubles a number, then cubes the result. If you input x, the machine does:
Inner function:u=2x (doubling)
Outer function:y=u3 (cubing)
Now, how fast does the final output change when you change x? It's not just the rate of the outer function, nor just the inner. It's the product of the two rates:
The outer function changes at rate dudy=3u2 (with respect to its own input u).
The inner function changes at rate dxdu=2 (with respect to x).
So the overall rate is:
dxdy=dudy⋅dxdu=3u2⋅2=6u2=6(2x)2=24x2
That's the chain rule: differentiate the outside, keep the inside the same, then multiply by the derivative of the inside.
dxdf(g(x))=f′(g(x))⋅g′(x)
The Precise Statement
If y=f(u) and u=g(x), then y is a function of x through u. The derivative is:
dxdy=dudy⋅dxdu
In function notation, if h(x)=f(g(x)), then:
h′(x)=f′(g(x))⋅g′(x)
The key: you evaluate the derivative of the outer function at the inner function (not at x), then multiply by the derivative of the inner function at x.
Step-by-Step Examples
Example 1: Differentiate y=(3x2+5)7
Outer function: (⋅)7, derivative: 7(⋅)6
Inner function: u=3x2+5, derivative: 6x
Apply the rule:
y′=7(3x2+5)6⋅(6x)=42x(3x2+5)6
Example 2: Differentiate y=1−x2
Rewrite: y=(1−x2)1/2
Outer: (⋅)1/2, derivative: 21(⋅)−1/2
Inner: u=1−x2, derivative: −2x
Apply:
y′=21(1−x2)−1/2⋅(−2x)=−1−x2x
Watch out
A common mistake: forgetting to multiply by the derivative of the inner function. For (3x2+5)7, some students write 7(3x2+5)6 and stop. That's wrong — you must also multiply by 6x.
Why It Works (Briefly)
The chain rule is a direct consequence of how limits work. If y changes by Δy when u changes by Δu, and u changes by Δu when x changes by Δx, then:
ΔxΔy=ΔuΔy⋅ΔxΔu
Taking limits as Δx→0 (which makes Δu→0 as well, provided g is continuous) gives the chain rule. The only subtlety: Δu might be zero for some small Δx, but for well-behaved functions this doesn't cause trouble.
The General Pattern for Algebraic Functions
For any algebraic function of the form [u(x)]n, the derivative is:
dxd[u(x)]n=n[u(x)]n−1⋅u′(x)
This is the power chain rule — it's the most common application. The same idea extends to any composition: trigonometric, exponential, logarithmic — the pattern is always the same.
Tip
When differentiating, first identify the "outer" and "inner" functions. Write them down explicitly if needed. Then: differentiate the outer, keep the inner untouched, multiply by the derivative of the inner. Practice this sequence until it becomes automatic.
The tangent line at any point x=x0 needs the slope f′(x0) together with the point (x0,f(x0)), substituted into y−y0=f′(x0)(x−x0).
✓Final answer
Tangent at x=−2: y=5 (horizontal, since f′(−2)=0). Tangent at x=1: y=9x−4 (slope 9).
Compute f′(x), evaluate slope and point at each given x, and write the tangent-line equation y−y0=f′(x0)(x−x0) — these are the lines to draw in GeoGebra.
Equation of the tangent to y=f(x) at x=x0:
y−f(x0)=f′(x0)(x−x0)
with f(x)=x3+3x2+1⟹f′(x)=3x2+6x.
Differentiate f(x).
f′(x)=dxd(x3+3x2+1)=3x2+6x
Tangent at x=−2. Find f(−2) and f′(−2):
f(−2)=(−2)3+3(−2)2+1=−8+12+1=5
f′(−2)=3(−2)2+6(−2)=12−12=0
Tangent-line equation: y−5=0⋅(x−(−2)), i.e.
y=5(a horizontal line — the curve has a turning point near here)
Tangent at x=1. Find f(1) and f′(1):
f(1)=13+3(1)2+1=1+3+1=5
f′(1)=3(1)2+6(1)=3+6=9
Tangent-line equation: y−5=9(x−1)
y=9x−9+5=9x−4
To draw in GeoGebra: type f(x)=x^3+3x^2+1, then Tangent((-2,f(-2)),f) and Tangent((1,f(1)),f) (or directly type y=5 and y=9x-4) to overlay both tangent lines on the curve.
Self-check. At x=−2: since f′(−2)=0, the curve momentarily flattens — matches a local extremum there (in fact f has a local max at x=−2, since f′ changes sign). At x=1: substituting back, the tangent line gives y=9(1)−4=5=f(1). ✓ Point lies on the tangent, as required.