Derivatives of Some Standard Composite Functions (Table 1.1.2)

Applying the chain rule to every entry of the Class 11 table, but with a general differentiable inner function $f(x)$ in place of plain $x$, produces the composite-function reference table used throughout this chapter (every entry carries an extra factor of $f'(x)$ compared to the plain table):

When a solution introduces $u=f(x)$ explicitly and applies the chain rule as two separate steps, that is called Method 1; once the pattern is familiar, it is faster to mentally 'treat the inner block as u' and write the derivative in one line — that is Method 2, used for the rest of the chapter.

Worked Example 1 — differentiate w.r.t. x:

1. $y=\sqrt{x^2+5}$: with $u=x^2+5$, $\dfrac{dy}{du}=\dfrac1{2\sqrt u}$ and $\dfrac{du}{dx}=2x$, so $\dfrac{dy}{dx}=\dfrac{2x}{2\sqrt{x^2+5}}=\dfrac{x}{\sqrt{x^2+5}}$.
2. $y=\sin(\log x)$: $\dfrac{dy}{dx}=\cos(\log x)\cdot\dfrac1x=\dfrac{\cos(\log x)}{x}$.
3. $y=e^{\tan x}$: $\dfrac{dy}{dx}=e^{\tan x}\sec^2x$.
4. $y=\log(x^5+4)$: $\dfrac{dy}{dx}=\dfrac{5x^4}{x^5+4}$.
5. $y=5^{3\cos x-2}$: $\dfrac{dy}{dx}=5^{3\cos x-2}\log5\cdot(-3\sin x)=-3\sin x\cdot5^{3\cos x-2}\log5$.
6. $y=\sqrt[3]{(2x^2-7)^5}=(2x^2-7)^{5/3}$: $\dfrac{dy}{dx}=\dfrac53(2x^2-7)^{2/3}\cdot4x=\dfrac{20x}{3}(2x^2-7)^{2/3}$. Worked Example 2 — differentiate w.r.t. x:

(i) $y=\sqrt{\sin x^3}$: $\dfrac{dy}{dx}=\dfrac{\cos(x^3)\cdot3x^2}{2\sqrt{\sin x^3}}=\dfrac{3x^2\cos(x^3)}{2\sqrt{\sin x^3}}$.

(ii) $y=\cot^2(x^3)$: $\dfrac{dy}{dx}=2\cot(x^3)\cdot[-\text{cosec}^2(x^3)]\cdot3x^2=-6x^2\cot(x^3)\text{cosec}^2(x^3)$.

(iii) $y=\log[\cos(x^5)]$: $\dfrac{dy}{dx}=\dfrac{-\sin(x^5)\cdot5x^4}{\cos(x^5)}=-5x^4\tan(x^5)$.

(iv) $y=(x^3+2x-3)^4(x+\cos x)^3$: by the product rule, each factor is differentiated in turn (using the chain rule inside each): $\dfrac{dy}{dx}=3(x^3+2x-3)^4(x+\cos x)^2(1-\sin x)+4(3x^2+2)(x^3+2x-3)^3(x+\cos x)^3$.

(v) $y=(1+\cos^2x)^4\sqrt x+\sqrt{\tan x}$: differentiate the product term and the root term separately. $\dfrac{d}{dx}\!\left[(1+\cos^2x)^4\right]=4(1+\cos^2x)^3\cdot2\cos x(-\sin x)=-8(1+\cos^2x)^3\sin x\cos x$, so by the product rule $\dfrac{d}{dx}\!\left[(1+\cos^2x)^4\sqrt x\right]=\dfrac{(1+\cos^2x)^4}{2\sqrt x}-8\sqrt x(1+\cos^2x)^3\sin x\cos x$; and $\dfrac{d}{dx}\sqrt{\tan x}=\dfrac{\sec^2x}{2\sqrt{\tan x}}$. Adding, $\dfrac{dy}{dx}=\dfrac{(1+\cos^2x)^4}{2\sqrt x}-8\sqrt x(1+\cos^2x)^3\sin x\cos x+\dfrac{\sec^2x}{2\sqrt{\tan x}}.$

Worked Example 3 — differentiate w.r.t. x (composites of a logarithm to a non-$e$ base, handled via the change-of-base formula $\log_a u=\dfrac{\ln u}{\ln a}$, so a constant factor $\dfrac1{\ln a}$ simply carries through the derivative):

(i) $y=\log_3(\log_5x)$. Writing $\log_5x=\dfrac{\ln x}{\ln5}$ so $y=\log_3(\ln x)-\log_3(\ln5)$, and the second term is a constant, $\dfrac{dy}{dx}=\dfrac{1}{\ln3}\cdot\dfrac{1}{\ln x}\cdot\dfrac1x=\dfrac1{x\ln x\ln3}$.

Parts (ii)–(iv) of this example continue in the same spirit — a logarithm of a logarithm, or a logarithm of a trigonometric/algebraic composite, to a non-standard base — and are differentiated the same way: rewrite with the change-of-base formula so the base becomes a constant multiplier, then apply the ordinary chain rule to what remains; the source scan for these particular sub-parts was too fragmented to transcribe safely, so their exact printed expressions are not reproduced here, but the technique demonstrated in parts (i) and (v)–(vi) is the same one that resolves them.

(v) A composite built from $a^{\log_a f(x)}=f(x)$ collapses first: once simplified, this example reduces to $y=\sin^2x+\cos^2x=1$, a constant, so $\dfrac{dy}{dx}=0$ — a reminder to always simplify algebraically/trigonometrically before differentiating, since a constant function's derivative is trivially zero no matter how complicated it looked originally.

(vi) $y=a^{\cot x}$ (again using $a^{\log_a f(x)}=f(x)$ to arrive at this simplified form first): $\dfrac{dy}{dx}=a^{\cot x}\log a\cdot(-\text{cosec}^2x)=-\text{cosec}^2x\cdot a^{\cot x}\log a$. …