Chain, product and implicit differentiation

What this concept is

For differentiable u,v, (uv)′=u′v+uv′ and (u/v)′=(u′v−uv′)/v² when v≠0. For a composition, (f(g(x)))′=f′(g(x))g′(x).

Why it matters here

Nested models are common: forgetting the inner derivative gives the wrong sensitivity even if the outer derivative is right.

Concrete example

For f(x)=x exp(−2x), product and chain rules give f′(x)=exp(−2x)(1−2x). For ln(1+x²), f′(x)=2x/(1+x²). Along x²+y²=25, implicit differentiation gives 2x+2y y′=0, so y′=−x/y where y≠0.

Common confusion / misconception

The derivative of a product is not the product of derivatives. Implicit differentiation treats y as depending on x; a vertical tangent may lack a finite y′.

How to use it

Use d(exp x)/dx=exp x, d(ln x)/dx=1/x for x>0, d(sin x)/dx=cos x, d(cos x)/dx=−sin x with radians. Identify the outermost operation before applying a rule. Differentiate exp(3x²) as exp(3x²)·6x; name each factor. Check one numerical slope only as a diagnostic, not proof.

Check yourself

Find d(sin(2x))/dx and explain why the factor 2 is necessary.

Mini drill or application

Differentiate x² ln x on x>0 and (x+1)/(x−1) on x≠1. Answers: 2x ln x+x and −2/(x−1)². Derive these, then invent a changed example with a different inner function.

Read this only if stuck

OpenStax selective support. Use the reading guide for the section sequence and practice contract. Record your reasoning before consulting solutions.