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NCERT Exemplar · Q26

Q.Differentiate w.r.t. xx: 8xx8\dfrac{8^x}{x^8}.

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We differentiate 8xx8\frac{8^x}{x^8} using the quotient rule, but because both numerator and denominator are non‑standard functions (exponential and power), we first rewrite using logarithms or apply the quotient rule directly with careful derivative formulas. The final derivative is 8x(xlog⁡8−8)x9\frac{8^x (x \log 8 - 8)}{x^9}.

The problem asks us to differentiate 8xx8\frac{8^x}{x^8} with respect to xx. At first glance, this looks like a straightforward quotient rule problem. But there’s a subtlety: the numerator 8x8^x is an exponential function (base constant, exponent variable), while the denominator x8x^8 is a power function (base variable, exponent constant). Their derivatives are different in form, and mixing them in a quotient requires care.

The key idea is to apply the quotient rule:

ddx(uv)=u′v−uv′v2\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}

where u=8xu = 8^x and v=x8v = x^8. Then we need u′u' and v′v' correctly.

Watch out

A common mistake is to treat 8x8^x as if it were x8x^8 and write its derivative as 8x78x^{7}. That is wrong — the derivative of axa^x (with aa constant) is axlog⁡aa^x \log a, not xax−1x a^{x-1}. The power rule only applies when the variable is in the base, not the exponent.

Let’s proceed step by step.

  1. Identify uu and vv

    Let u=8xu = 8^x and v=x8v = x^8.

  2. Differentiate u=8xu = 8^x

    The derivative of an exponential axa^x is axlog⁡aa^x \log a. So:

u′=8xlog⁡8.u' = 8^x \log 8.

No exponent reduction — just multiply by the natural log of the base.

  1. Differentiate v=x8v = x^8 This is a standard power rule: bring down the exponent 8, reduce the exponent by 1:

v′=8x7.v' = 8x^7.

  1. Apply the quotient rule

ddx(8xx8)=(8xlog⁡8)⋅x8−8x⋅(8x7)(x8)2.\frac{d}{dx}\left(\frac{8^x}{x^8}\right) = \frac{(8^x \log 8) \cdot x^8 - 8^x \cdot (8x^7)}{(x^8)^2}.

  1. Simplify the numerator Factor out the common term 8x8^x: Numerator=8x(x8log⁡8−8x7).\text{Numerator} = 8^x \left( x^8 \log 8 - 8x^7 \right). …

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