Question

find the differentiation of s=5t^3-3t^5​

Answer 1

$\frac{d}{dx}x^n=nx^{n-1}$

So

$\frac{ds}{dt}=3\times 5t^2-3\times 5t^4\\\\\implies \frac{ds}{dt}=15t^2+15t^4\\\\\implies 15t^2(1+t^2)$

Answer 2

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Calculus

Find ds/dt s=5t^3-3t^5

s=5t3−3t5s=5t3-3t5

Differentiate both sides of the equation.

ddt(s)=ddt(5t3−3t5)ddt(s)=ddt(5t3-3t5)

The derivative of ss with respect to tt is s's′.

s's′

Differentiate the right side of the equation.

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By the Sum Rule, the derivative of 5t3−3t55t3-3t5 with respect to tt is ddt[5t3]+ddt[−3t5]ddt[5t3]+ddt[-3t5].

ddt[5t3]+ddt[−3t5]ddt[5t3]+ddt[-3t5]

Evaluate ddt[5t3]ddt[5t3].

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15t2+ddt[−3t5]15t2+ddt[-3t5]

Evaluate ddt[−3t5]ddt[-3t5].

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15t2−15t415t2-15t4

Reorder terms.

−15t4+15t2