Differentiating D(x) = cos(sin(2x)) requires nested chain rule. Outer cos(g), derivative -sin(g), g = sin(2x). g' = cos(2x) * 2. So D'(x) = -sin(g) * 2 cos(2x). A common omission is forgetting the 2 from the inner argument. Pattern: Spot double trig compositions and apply chain twice, outer to inner.

Cosine of a power function requires multiple chain rule applications. Outer $\cos(u)$, $u = (x^2 - 1)^4$, derivative $-\sin(u)$, inner $u' = 4(x^2 - 1)^3 \cdot 2x = 8x(x^2 - 1)^3$. $y'(x) = -\sin((x^2 - 1)^4) \cdot 8x(x^2 - 1)^3$. Common omission: forgetting the 2x from inside the power. Missing the 4 from exponent or negative sign. Recognize trig of power of quadratic and layer the chain rule. This strategy applies to deep nests, ensuring thorough derivatives.

This tangent of a log requires chain rule for trig-log composite. Outer tan(u), u = $ln(x^2$ + 4), derivative $sec^2$(u), inner u' = $2x/(x^2$ + 4). So u'(x) = $(2x/(x^2$ + 4)) $sec^2$$(ln(x^2$ + 4)). Common omission: forgetting 2x from inner derivative. Confusing tan derivative with sec tan is another error. Recognize trig of log polynomial and chain fully. This pattern aids in complex nests, building derivative skills.

Differentiating log of square root uses chain rule after simplification. $v(x) = \frac{1}{2} \ln(1 - \sin x)$, $v'(x) = \frac{1}{2} \cdot \frac{-\cos x}{1 - \sin x}$. Outer ln, inner $\sqrt{1 - \sin x}$, but simplify first. Common omission: missing the 1/2 factor. Forgetting negative from -sin x derivative. Spot log of sqrt trig and use properties then chain. This transfers to radical logs, improving accuracy.

For G(x) = tan(1 / ln x), chain rule applies. Outer tan(j), derivative sec²(j), j = (ln $x)^{-1}$. j' = - (ln $x)^{-2}$ * (1/x) = -1/(x (ln x)²). Thus, G'(x) = sec²(j) * (-1/(x (ln x)²)). A common omission is sign error in the reciprocal derivative. Recognize trig of inverse log and use power rule in chain.

This is ln composed with sin(x³). The derivative of ln(u) is 1/u, giving us 1/sin(x³) times the derivative of sin(x³). The derivative of sin(x³) requires another application of the chain rule: cos(x³) · 3x². Combining these: q'(x) = (1/sin(x³)) · cos(x³) · 3x² = 3x²cos(x³)/sin(x³). This can also be written as 3x²cot(x³). A common error is forgetting the 3x² from differentiating x³. Pattern tip: ln(sin(power function)) derivatives always include the power function's derivative as a factor in the numerator.

Differentiating H(x) = cos(ln((x² + 1)/x)) uses chain. Outer cos(k), derivative -sin(k), k = ln((x² + 1)/x) = ln(x² + 1) - ln x. k' = (2x/(x² + 1)) - 1/x = (x² - 1)/(x (x² + 1)). So H'(x) = -sin(k) * (x² - 1)/(x (x² + 1)). A common omission is wrong sign in k'. Pattern: Simplify log quotients before chaining with outer trig.

To find the derivative of f(t) = ln(5 - 2cos(3t)), we apply the chain rule with a logarithmic outer function and a trigonometric expression inside. The derivative of ln(u) is 1/u, so we get 1/(5 - 2cos(3t)) as the outer derivative. For the inner function 5 - 2cos(3t), we differentiate to get -2(-sin(3t))·3 = 6sin(3t), where we must remember both the chain rule for cos(3t) and the coefficient -2. Multiplying outer and inner derivatives gives f'(t) = [1/(5 - 2cos(3t))]·6sin(3t) = 6sin(3t)/(5 - 2cos(3t)). A common error is forgetting to multiply by 3 when differentiating cos(3t). When you see ln(trig expression), always identify the outer logarithm and inner trigonometric function, then multiply their derivatives systematically.

We have tan composed with ln(x²+1). The derivative of tan(u) is sec²(u), so we get sec²(ln(x²+1)) times the derivative of ln(x²+1). The derivative of ln(x²+1) is 1/(x²+1) times 2x (from differentiating x²+1). Therefore, u'(x) = sec²(ln(x²+1)) · 2x/(x²+1). A common error is forgetting the 2x from the derivative of x²+1 or confusing tan's derivative with sin or cos derivatives. Remember: tan(ln(polynomial)) produces sec² of the same argument, always with the polynomial's derivative in the numerator of the accompanying fraction.

This involves sec composed with (x-2)⁵. The derivative of sec(u) is sec(u)tan(u), and we multiply by the derivative of (x-2)⁵. Using the chain rule: d/dx[(x-2)⁵] = 5(x-2)⁴. Therefore, s'(x) = sec((x-2)⁵)tan((x-2)⁵) · 5(x-2)⁴. A common mistake is confusing the derivative of sec with that of tan (which would give sec²). Remember the mnemonic: the derivative of sec involves both sec and tan. Pattern: sec(power function) derivatives always include the original sec, tan of the same argument, and the power rule applied to the inner function.