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# Find the derivative of $10\sqrt{x^2+25}-50$

## Step-by-step Solution

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###  Videos

$\frac{10x}{\sqrt{x^2+25}}$
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##  Step-by-step Solution 

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The derivative of a sum of two or more functions is the sum of the derivatives of each function

$\frac{d}{dx}\left(10\sqrt{x^2+25}\right)+\frac{d}{dx}\left(-50\right)$

Learn how to solve differential calculus problems step by step online.

$\frac{d}{dx}\left(10\sqrt{x^2+25}\right)+\frac{d}{dx}\left(-50\right)$

Learn how to solve differential calculus problems step by step online. Find the derivative of 10(x^2+25)^1/2-50. The derivative of a sum of two or more functions is the sum of the derivatives of each function. The derivative of the constant function (-50) is equal to zero. The derivative of a function multiplied by a constant (10) is equal to the constant times the derivative of the function. The power rule for differentiation states that if n is a real number and f(x) = x^n, then f'(x) = nx^{n-1}.

$\frac{10x}{\sqrt{x^2+25}}$

##  Explore different ways to solve this problem

Solving a math problem using different methods is important because it enhances understanding, encourages critical thinking, allows for multiple solutions, and develops problem-solving strategies. Read more

Find the derivativeFind derivative of (10(x^2+25)^0.5-50) using the product ruleFind derivative of (10(x^2+25)^0.5-50) using the quotient ruleFind derivative of (10(x^2+25)^0.5-50) using logarithmic differentiationFind derivative of (10(x^2+25)^0.5-50) using the definition

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4
5
6
7
8
9
0
a
b
c
d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
2

e
π
ln
log
log
lim
d/dx
Dx
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

### Main Topic: Differential Calculus

The derivative of a function of a real variable measures the sensitivity to change of a quantity (a function value or dependent variable) which is determined by another quantity (the independent variable). Derivatives are a fundamental tool of calculus.