Write this in the form ddx(3)=0{\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}(3)=0}. This says “The derivative of 3 with respect to x is 0. " The derivative of a term is the “rate of change” of that term: how quickly that term changes inside a function. Since a constant never changes (3 will always stay 3), its rate of change is always zero.
Write this as ddx(x)=1{\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}(x)=1}. The notation dx means “derivative with respect to x. " This means that we are changing the value of x, and seeing how much faster or slower the other term changes in response. In ddx(x){\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}(x)}, we’re comparing the change in x to the change in x. That’s the same thing, which is why the rate of change is 1.
Have you noticed a pattern? In the derivative, the value of the variable’s exponent is always one lower than it was in the original term. x2{\displaystyle x^{2}} gets “downgraded” to x1{\displaystyle x^{1}} (which is x), and x1{\displaystyle x^{1}} gets “downgraded” to x0{\displaystyle x^{0}} (which equals 1). Since the value of the variable’s exponent is called the “degree” of the polynomial, we can say that differentiating a term reduces the degree of that term by one. [5] X Research source
Example: what is ddx(x7){\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}(x^{7})} (the derivative of x7{\displaystyle x^{7}} with respect to x)? The exponent, 7, becomes a coefficient in front of the term: 7x?{\displaystyle 7x^{?}} The new exponent is one lower than the original, 7-1=6. The answer is 7x6{\displaystyle 7x^{6}}.
Example: what is ddx(5x3){\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}(5x^{3})}? ddx(5x3)=5×ddx(x3){\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}(5x^{3})=5\times {\frac {\mathrm {d} }{\mathrm {d} x}}(x^{3})} (This means we can find the derivative of x3{\displaystyle x^{3}}, then multiply our answer by 5. ) To find the derivative of x3{\displaystyle x^{3}}, make the exponent 3 a coefficient, then reduce the exponent by 1: ddx(x3)=3x2{\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}(x^{3})=3x^{2}} Plug this back into your formula and multiply the two coefficients together: 5×ddx(x3)=5×3x2=15x2{\displaystyle 5\times {\frac {\mathrm {d} }{\mathrm {d} x}}(x^{3})=5\times 3x^{2}=15x^{2}}
For example, take f(x)=−12x3+x2−5x+6{\displaystyle f(x)=-12x^{3}+x^{2}-5x+6}. The derivative, f′(x){\displaystyle f’(x)}, is equal to the derivative of each term, added or subtracted as they were in the original. In mathematical terms, we can write this as:ddx(f(x))=ddx(−12x3+x2−5x+6){\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}(f(x))={\frac {\mathrm {d} }{\mathrm {d} x}}(-12x^{3}+x^{2}-5x+6)} =ddx(−12x3)+ddx(x2)−ddx(5x)+ddx(6){\displaystyle ={\frac {\mathrm {d} }{\mathrm {d} x}}(-12x^{3})+{\frac {\mathrm {d} }{\mathrm {d} x}}(x^{2})-{\frac {\mathrm {d} }{\mathrm {d} x}}(5x)+{\frac {\mathrm {d} }{\mathrm {d} x}}(6)}.
In our example, 6 is the constant. ddx(6)=0{\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}(6)=0}, so we can get rid of it. Careful: only terms with no variables are constants. This rule does not affect numbers that are multiplied by x or any other variable.
f′(x)=ddx(−12x3)+ddx(x2)−ddx(5x){\displaystyle f’(x)={\frac {\mathrm {d} }{\mathrm {d} x}}(-12x^{3})+{\frac {\mathrm {d} }{\mathrm {d} x}}(x^{2})-{\frac {\mathrm {d} }{\mathrm {d} x}}(5x)} =(3×−12)x?+2x?−(1×5)x?{\displaystyle =(3\times -12)x^{?}+2x^{?}-(1\times 5)x^{?}} =−36x?+2x?−5x?{\displaystyle =-36x^{?}+2x^{?}-5x^{?}} Since x=x1{\displaystyle x=x^{1}}, we took the “1” exponent and moved it in front of the 5x{\displaystyle 5x} term. Since multiplying by 1 never changes the term, you can skip this step once you understand what’s going on.
f′(x)=−36x3−1+2x2−1−5x1−1{\displaystyle f’(x)=-36x^{3-1}+2x^{2-1}-5x^{1-1}} =−36x2+2x1−5x0{\displaystyle =-36x^{2}+2x^{1}-5x^{0}} =−36x2+2x−5{\displaystyle =-36x^{2}+2x-5} Remember that x1{\displaystyle x^{1}} is the same as x{\displaystyle x}. Also remember that anything raised to the zeroth power (x0{\displaystyle x^{0}}) equals 1.
For example, evaluate the derivative f’(x) at x=2. The derivative equation we found is f′(x)=−36x2+2x−5{\displaystyle f’(x)=-36x^{2}+2x-5} f′(2)=−36(2)2+2(2)−5{\displaystyle f’(2)=-36(2)^{2}+2(2)-5} =−36(4)+4−5{\displaystyle =-36(4)+4-5} =−144+4−5{\displaystyle =-144+4-5} =−145{\displaystyle =-145} This answer relates back to the original function f(x). It tells us that, if we draw a tangent line to that function at x=2, the slope of that tangent line is -145.
