Advertisements
Advertisements
Question
Find the derivative of x5 (3 – 6x–9).
Advertisements
Solution
Let f (x) = x5 (3 – 6x–9)
By Leibnitz product rule,
f'(x) = `x^5 d/(dx) (3 - 6x^-9) + (3 - 6x^-9) d/(dx) (x^5)`
= x5 {0 - 6(-9)x-9-1} + (3 - 6x-9)(5x4)
= x5 (54x-10) + 15x4 - 30x-5
= 54x-5 + 15x4 - 30x-5
= 24x-5 + 15x4
= `15x^4 + 24/x^5`
APPEARS IN
RELATED QUESTIONS
Find the derivative of the following function (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers):
`(1 + 1/x)/(1- 1/x)`
\[\frac{2}{x}\]
\[\frac{x^2 + 1}{x}\]
k xn
Differentiate of the following from first principle:
− x
Differentiate of the following from first principle:
sin (2x − 3)
Differentiate each of the following from first principle:
\[\frac{\cos x}{x}\]
Differentiate each of the following from first principle:
x2 sin x
Differentiate each of the following from first principle:
sin x + cos x
Differentiate each of the following from first principle:
\[e^\sqrt{2x}\]
tan2 x
tan (2x + 1)
\[\sqrt{\tan x}\]
\[\cos \sqrt{x}\]
\[\tan \sqrt{x}\]
\[\tan \sqrt{x}\]
x4 − 2 sin x + 3 cos x
ex log a + ea long x + ea log a
\[\frac{2 x^2 + 3x + 4}{x}\]
\[\frac{a \cos x + b \sin x + c}{\sin x}\]
x3 ex
sin x cos x
\[\frac{2^x \cot x}{\sqrt{x}}\]
\[\frac{x^2 \cos\frac{\pi}{4}}{\sin x}\]
x4 (5 sin x − 3 cos x)
Differentiate each of the following functions by the product rule and the other method and verify that answer from both the methods is the same.
(3x2 + 2)2
\[\frac{x^2 + 1}{x + 1}\]
\[\frac{x^2 - x + 1}{x^2 + x + 1}\]
\[\frac{a + \sin x}{1 + a \sin x}\]
\[\frac{1 + 3^x}{1 - 3^x}\]
\[\frac{p x^2 + qx + r}{ax + b}\]
\[\frac{x + \cos x}{\tan x}\]
\[\frac{x}{\sin^n x}\]
Write the value of \[\frac{d}{dx} \left\{ \left( x + \left| x \right| \right) \left| x \right| \right\}\]
Mark the correct alternative in of the following:
If \[y = \sqrt{x} + \frac{1}{\sqrt{x}}\] then \[\frac{dy}{dx}\] at x = 1 is
Find the derivative of 2x4 + x.
(ax2 + cot x)(p + q cos x)
