Q.Differentiate $y = 6x^{5} - 4e^{x} + 3\ln x - 2(5^{x}) + 11$ with respect to $x$.

Differentiate term by term using the power, exponential, logarithmic, general-exponential, and constant rules.\n> [!TLDR]\n> Apply each of the five standard formulae to its own term.\n\n$\dfrac{dy}{dx} = 30x^4 - 4e^x + \dfrac{3}{x …

Differentiate each term using §7's standard formulae, term by term (valid by the sum/difference/constant-multiple property of differentiation):

• $\dfrac{d}{dx}(6x^5) = 6\cdot5x^{4} = 30x^{4}$ (power rule, constant-multiple $6$).
• $\dfrac{d}{dx}(-4e^x) = -4e^{x}$ (exponential rule, constant-multiple $-4$; $e^x$ is its own derivative).
• $\dfrac{d}{dx}(3\ln x) = 3\cdot\dfrac1x = \dfrac{3}{x}$ (logarithmic rule, constant-multiple $3$).
• $\dfrac{d}{dx}\big(-2(5^x)\big) = -2\cdot 5^{x}\ln5$ (general exponential rule $a^x\to a^x\ln a$ with $a=5$, constant-multiple $-2$).
• $\dfrac{d}{dx}(11) = 0$ (constant rule).

Adding the pieces:

$\frac{dy}{dx} = 30x^{4} - 4e^{x} + \frac{3}{x} - 2(5^{x})\ln5 + 0 = 30x^{4} - 4e^{x} + \frac{3}{x} - 2(5^{x})\ln5.$ …

A numeric difference-quotient check at $x=1$ (comparing $\frac{y(1.001)-y(0.999)}{0.002}$ against the formula's value at $x=1$) gives closely matching res …

The most common slip on this type of question is applying the power rule's exponent-times-coefficient pattern to the $a^x$ term too — writing $\frac{d}{dx}(5^x)$ as $x\cdot5^{x-1}$ (the power-rule pattern for $x^n$) instead of the correct general-exponen …