y = 4x^3 - 5/x^2 find dy/dx
First rewrite the equation into an easier form:y = y = 4x^3 - 5x^-2dy/dx means differentiate y with respect to xSo when differentiating, you look at the first term and write the co-efficient of the x term ( so in this case the first term is 4x^3 and the co-efficient is 4) then bring down the value of the exponent and multiply it to the co-efficient (hence 3 is the exponent). Then decrease the value of the exponent by 1. Do this for all the terms in the equation. dy/dx = (4)(3)x^2 - (5)(-2)x^-3If we simplify this equation we get:dy/dx = 12x^2 +10x^-3 ( remember 2 negatives multiplied together give a positive)now lets put the equation back tothe form it was given in the question (i.e. with the fraction in the second term)dy/dx = 12x^2 +10/x^3
Related Maths A Level answers
All answers ▸A curve has parametric equations x=t(t-1), y=4t/(1-t). The point S on the curve has parameter t=-1. Show that the tangent to the curve at S has equation x+3y+4=0.
f(x)=(2x+1)/(x-1) with domain x>3. (a)Find the inverse of f(x). (b)Find the range of f(x). (c) g(x)=x+5 for all x. Find the value of x such that fg(x)=3.
If the functions f and g are defined: f: x--> x/5 + 4 g : x--> 30x + 10. what is x, if fg(x) = x. ?? What would fgf(x) = x^2 be??
Show that the integral of tan(x) is ln|sec(x)| + C where C is a constant.